Inequalities Marathon

Inequalities Marathon

Problem 1 ‘India 2002’ (Hassan Al-Sibyani): For any positive real numbers a, b, c show that the following inequality holds a b c c+a a+b b+c + + ≥ + + b c a c+b a+c b+a

First Solution (Popa Alexandru): Ok. After not so many computations i got that: a b c a+b b+c c+a + + − − − b c a c+a a+b b+c  2  b2 c2 a b abc a c + 2+ 2− − − = (a + b)(b + c)(c + a) b2 c a b c a   abc bc ca ab + + + − 3 (a + b)(b + c)(c + a) c2 a2 b2 a2 b2 c2 a b c ab bc ca So in order to prove the above inequality we need to prove 2 + 2 + 2 ≥ + + and 2 + 2 + 2 ≥ 3 b c a b c a c a b The second inequality is obvious by AM-GM , and the for the first we have: 

b2 c2 a2 + 2+ 2 2 b c a

2

 ≥3

a2 b2 c2 + 2+ 2 2 b c a



 ≥

c a b + + b c a

2

a b c where i used AM-GM and the inequality 3(x2 + y 2 + z 2 ) ≥ (x + y + z)2 for x = , y = , z = b c a So the inequality is proved. a b c Second Solution (Raghav Grover): Substitute = x, = y, = z So xyz = 1. The inequality after b c a substitution becomes x2 z + y 2 x + z 2 x + x2 + y 2 + z 2 ≥ x + y + z + 3 x2 z + y 2 x + z 2 x ≥ 3 So now it is left to prove that x2 + y 2 + z 2 ≥ x + y + z which is easy. Third Solution (Popa Alexandru): Bashing out it gives a4 c2 + b4 a2 + c4 b2 + a3 b3 + b3 c3 + a3 c3 ≥ abc(ab2 + bc2 + ca2 + 3abc) which is true because AM-GM gives : a3 b3 + b3 c3 + a3 c3 ≥ 3a2 b2 c2 and by Muirhead : a4 c2 + b4 a2 + c4 b2 ≥ abc(ab2 + +bc2 + ca2 )

Fourth Solution (Popa Alexandru): Observe that the inequality is equivalent with: X a2 + bc ≥3 a(a + b) cyc Now use AM-GM:

s Q X a2 + bc (a2 + bc) Q ≥33 a(a + b) abc (a + b) cyc

2

So it remains to prove: Y Y (a2 + bc) ≥ abc (a + b) Now we prove (a2 + bc)(b2 + ca) ≥ ab(c + a)(b + c) ⇔ a3 + b3 ≥ ab2 + a2 b ⇔ (a + b)(a − b)2 ≥ 0 Multiplying the similars we are done.

Problem 2 ‘Maxim Bogdan’ (Popa Alexandru): Let a, b, c, d > 0 such that a ≤ b ≤ c ≤ d and abcd = 1 . Then show that: 3 (a + 1)(d + 1) ≥ 3 + 3 4d 1 First Solution (Mateescu Constantin):From the condition a ≤ b ≤ c ≤ d we get that a ≥ 3 . d   1 =⇒ (a + 1)(d + 1) ≥ + 1 (d + 1) 3  d  1 3 Now let’s prove that 1 + 3 (d + 1) ≥ 3 + 3 d 4d 3 This is equivalent with: (d3 + 1)(d + 1) ≥ 3d3 + 4  2 3 2 ⇐⇒ [d(d − 1)] − [d(d − 1)] + 1 ≥ ⇐⇒ d(d − 1) − 21 ≥ 0. 4 √ 1 1 1+ 3 Equality holds for a = 3 and d(d − 1) − = 0 ⇐⇒ d = d 2 2

Problem 3 ‘Darij Grinberg’ (Hassan Al-Sibyani): If a, b, c are three positive real numbers, then a 2

(b + c)

+

b (c + a)

2

+

c (a + b)

2

≥

9 4 (a + b + c)

First Solution (Dimitris X): X (a + b + c)2 a2 P ≥ 2 ab2 + ac2 + 2abc sym a b + 6abc So we only have to prove X that: X X 4(a + b + c)3 ≥ 9 a2 b + 54abc ⇐⇒ 4(a3 + b3 + c3 ) + 12 a2 b + 24abc ≥ 9 a2 b + 54abc ⇐⇒ 4(a3 + b3 + c3 ) + 3

sym X a2 b ≥ 30abc

sym

sym

sym

But

X

2

a b ≥ 6abc and a3 + b3 + c3 ≥ 3abc

sym

So 4(a3 + b3 + c3 ) + 3

X

a2 b ≥ 30abc

sym

Second Solution (Popa Alexandru): Use Cauchy-Schwartz and Nesbitt:    2 a b c a b c 9 (a + b + c) + + ≥ + + ≥ (b + c)2 (c + a)2 (a + b)2 b+c c+a a+b 4

3

Problem 4 ‘United Kingdom’ (Dimitris X): For a, b, c ≥ 0 and a+b+c = 1 prove that 7(ab+bc+ca) ≤ 2+9abc First Solution (Popa Alexandru): Homogenize to 2(a + b + c)3 + 9abc ≥ 7(ab + bc + ca)(a + b + c) Expanding it becomes : X

a3 + 6

sym

X

a2 b + 21abc ≥ 7

sym

X

a2 b + 21abc

sym

So we just need to show: X

a3 ≥

sym

X

a2 b

sym

which is obvious by a3 + a3 + b3 ≥ 3a2 b and similars. Second Solution (Popa Alexandru): Schur gives 1 + 9abc ≥ 4(ab + bc + ca) and use also 3(ab + bc + ca) ≤ (a + b + c)2 = 1 Suming is done .

Problem 5 ‘Gheorghe Szollosy, Gazeta Matematica’ (Popa Alexandru): Let x, y, z ∈ R+ . Prove that: p

x(y + 1) +

p p 3p y(z + 1) + z(x + 1) ≤ (x + 1)(y + 1)(z + 1) 2

First Solution (Endrit Fejzullahu): Dividing with the square root on the RHS we have : r r r 3 x y z + + ≤ (x + 1)(z + 1) (x + 1)(y + 1) (y + 1)(z + 1) 2 By AM-GM r

1 x ≤ (x + 1)(z + 1) 2



1 x + x+1 y+1



r

y 1 ≤ (x + 1)(y + 1) 2



y 1 + y+1 x+1



r

z 1 ≤ (y + 1)(z + 1) 2



z 1 + z+1 y+1



Summing we obtain 1 LHS ≤ 2



x 1 + x+1 x+1



 +

y 1 + y+1 y+1



 +

z 1 + z+1 z+1



Problem 6 ‘—’ (Endrit Fejzullahu): Let a, b, c be positive numbers , then prove that 1 1 1 4a 4b 4c + + ≥ 2 + 2 + 2 2 2 2 2 a b c 2a + b + c a + 2b + c a + b2 + 2c2

4

=

3 2

√ First Solution (Mateescu Constantin): By AM − GM we have 2a2 + b2 + c2 ≥ 4a bc 4a 1 4a ≤ √ =√ =⇒ 2a2 + b2 + c2 4a bc bc 1 1 1 Addind the similar inequalities =⇒ RHS ≤ √ + √ + √ (1) ca ab bc  2 2  1 1 1 1 1 1 Using Cauchy-Schwarz we have √ + √ + √ ≤ + + a b c ca ab bc 1 1 1 1 1 1 So √ + √ + √ ≤ + + (2) a b c ca ab bc From (1), (2) we obtain the desired result . Second Solution (Popa Alexandru): By Cauchy-Schwatz : a a 4a ≤ 2 + 2 2a2 + b2 + c2 a + b2 a + c2 Then we have RHS ≤

 X 1 X a+b X 2 1 ≤ ≤ + = LHS a2 + b2 a+b 2a 2b cyc cyc cyc

Problem 7 ‘—’ (Mateescu Constantin): Let a, b, c, d, e be non-negative real numbers such that a + b + c + d + e = 5 . Prove that: abc + bcd + cde + dea + eab ≤ 5

First Solution (Popa Alexandru): Assume e ≤ min{a, b, c, d}. Then AM-GM gives : e(c + a)(b + d) + bc(a + d − e) ≤

e(5 − e)2 (5 − 2e)2 + ≤5 4 27

the last one being equivalent with: (e − 1)2 (e + 8) ≥ 0

Problem 8 ‘Popa Alexandru’ (Popa Alexandru): Let a, b, c be real numbers such that 0 ≤ a ≤ b ≤ c . Prove that: (a + b)(c + a)2 ≥ 6abc

First Solution (Popa Alexandru): Let b = xa , c = yb = xya ⇒ x, y ≥ 1 Then:

(a + b)(a + c)2 ≥ 2abc 3 ⇔ (x + 1)(xy + 1)2 · a3 ≥ 6x2 ya3

5

⇔ (x + 1)(xy + 1)2 ≥ 6x2 y ⇔ (x + 1)(4xy + (xy − 1)2 ) ≥ 6x2 y ⇔ 4xy + (xy − 1)2 · x + (xy − 1)2 − 2x2 y ≥ 0 We have that: 4xy + (xy − 1)2 · x + (xy − 1)2 − 2x2 y ≥ ≥ 4xy + 2(xy − 1)2 − 2x2 y( because x ≥ 1) = 2x2 y 2 + 2 − 2x2 y = 2xy(y − 1) + 2 > 0 done. Second Solution (Endrit Fejzullahu): Let b = a + x, c = b + y = a + x + y,sure x, y ≥ 0 Inequality becomes (2a + x)(x + y + 2a)2 − 6a(a + x)(a + x + y) ≥ 0 But (2a + x)(x + y + 2a)2 − 6a(a + x)(a + x + y) = 2a3 + 2a2 y + 2axy + 2ay 2 + x3 + 2x2 y + xy 2 which is clearly positive.

Problem 9 ‘—’ (Raghav Grover): Prove for positive reals b c d a + + + ≥2 b+c c+d d+a a+b

First Solution (Dimitris X): (a + b + c + d)2 From andreescu LHS ≥ X ab + (ac + bd) sym

So we only need to prove that: (a + b + c + d)2 ≥ 2

X

ab + 2(ac + bd) ⇐⇒ (a − c)2 + (b − d)2 ≥ 0....

sym

Problem 10 ‘—’ (Dimitris X): Let a, b, c, d be REAL numbers such that a2 + b2 + c2 + d2 = 4 Prove that: a3 + b3 + c3 + d3 ≤ 8

First Solution (Popa Alexandru): Just observe that a3 + b3 + c3 + d3 ≤ 2(a2 + b2 + c2 + d2 ) = 8 because a, b, c, d ≤ 2

6

Problem 11 ‘—’ (Endrit Fejzullahu): Let a, b, c be positive real numbers such that abc = 1 .Prove that X cyc

1 1 ≤ a2 + 2b2 + 3 2

First Solution (Popa Alexandru): Using AM-GM we have : LHS =

X cyc

X 1 1 ≤ (a2 + b2 ) + (b2 + 1) + 2 2ab + 2b + 2 cyc =

because

1X 1 1 = 2 cyc ab + b + 1 2

1 1 1 ab 1 = = · = bc + c + 1 bc + c + abc c ab + b + 1 ab + b + 1

and

1 = ca + a + 1

so X cyc

1 b

1 b = ab + b + 1 +a+1

1 1 ab b = + + =1 ab + b + 1 ab + b + 1 ab + b + 1 ab + b + 1

Problem 12 ‘Popa Alexandru’ (Popa Alexandru): Let a, b, c > 0 such that a + b + c = 1. Prove that: 15 1+a+b 1+b+c 1+c+a + + ≥ 2+c 2+a 2+b 7

First Solution (Dimitris X): X1+a+b X 3 + (a + b + c) X 4 15 36 36 +1≥ + 3 ⇐⇒ ≥ ⇐⇒ ≥ 2+c 7 2+c 7 2+c 7 X 22 (2 + 2 + 2)2 36 But ≥ = 2+c 2+2+2+a+b+c 7 Second Solution (Endrit Fejzullahu): Let a ≥ b ≥ c then by Chebyshev’s inequality we have X 1 1 5X 1 LHS ≥ (1 + 1 + 1 + 2(a + b + c)) = 3 2+a 3 cyc 2 + a cyc X 1 9 15 By Titu’s Lemma ≥ , then LHS ≥ 2 + a 7 7 cyc

Problem 13 ‘Titu Andreescu, IMO 2000’ (Dimitris X): Let a, b, c be positive so that abc = 1     1 1 1 a−1+ b−1+ c−1+ ≤1 b c a

7

First Solution   (Endrit Fejzullahu):   1 1 1 a−1+ b−1+ c−1+ ≤1 b c a x y Substitute a = , b = y z Inequality isequivalent with  x z y x z y ≤1 −1+ −1+ −1+ y y z z x x ⇐⇒ (x + z − y)(y − z + x)(z − x + y) ≤ xyz WLOG ,Let x > y > z, then x + z > y, x + y > z.If y + z < x, then we are done because (x + z − y)(y − z + x)(z − x + y) ≤ 0 and xyz ≥ 0 Otherwise if y + z > x , then x, y, z are side lengths of a triangle ,and then we can make the substitution x = m + n, y = n + t and z = t + m Inequality is equivalent with 8mnt ≤ (m√+ n)(n + t)(t + √m), this is true by√AM-GM m + n ≥ 2 mn, n + t ≥ 2 nt and t + m ≥ 2 tm, multiply and we’re done.

Problem 14 ‘Korea 1998’ (Endrit Fejzullahu): Let a, b, c > 0 and a + b + c = abc. Prove that: √

1 a2

First Solution (Dimitris X): Setting a =

+1

+√

1 1 3 +√ ≤ 2 2 +1 c +1

b2

1 1 1 , b = , c = the condition becomes xy + yz + zx = 1, and the x y z

inequality: X x 3 √ ≤ 2 2 x +1 X Xr x X x x x √ p But = = 2 2 x+yx+z x +1 x + xy + xz + zy x x r + x x x+y x+z But ≤ x+yx+z 2 y x z y z x r + + + + + X x x 3 x+y x+y z+x z+x y+z z+y So ≤ = x+yx+z 2 2 Second Solution (Raghav Grover): Substitute a = tan x,b = tan y and c = tan z where x + y + z = π And we are left to prove cos x + cos y + cos z ≤ 32 Which i think is very well known.. √ Third Solution (Endrit Fejzullahu): By AM-GM we have a + b + c ≥ 3 3 abc and since a + b + c = abc =⇒ (abc)2 ≥ 27 We rewrite the given inequality  as 1 1 1 1 1 + + ≤ 3 a2 + 1 b2 + 1 c2 + 1 2

8

Since function f (a) = √

1

is concave ,we apply Jensen’s inequality +1    1 1 1 a+b+c abc 1 1 f (a) + f (b) + f (c) ≤ f =f =q ≤ ⇐⇒ (abc)2 ≥ 27, QED 2 3 3 3 3 3 2 (abc) +1 32 a2

Problem 15 ‘—’ (Dimitris X): If a, b, c ∈ R and a2 + b2 + c2 = 3. Find the minimum value of A = ab + bc + ca − 3(a + b + c). First Solution (Endrit Fejzullahu): (a + b + c)2 − 3 (a + b + c)2 − a2 − b2 − c2 = ab + bc + ca = 2 2 Let a + b + c = x 2 x 3 Then A = − 3x − 2 2 x2 3 We consider the second degree fuction f (x) = − 3x − 2 2   −b We obtain minimum for f = f (3) = −6 2a Then Amin = −6 ,it is attained for a = b = c = 1

Problem 16 ‘—’ (Endrit Fejzullahu): If a, b, c are positive real numbers such that a + b + c = 1.Prove that r a b c 3 √ +√ ≥ +√ 2 c+a b+c a+b x First Solution (keyree10): Let f (x) = √ . f ”(x) > 0 1−x X 1 3s a+b+c a √ = , by jensen’s. Therefore, ≥√ ,where s = 3 3 1−a 1−s r b 3 a c +√ ≥ . Hence proved =⇒ √ +√ 2 c+a b+c a+b Second Solution (geniusbliss): By  holders’ inequality,    X X X a a    a(b + c) ≥ (a + b + c)3 1 1 2 2 (b + c) (b + c) cyclic cyclic cyclic thus,  X  cyclic

2 a (b + c)

1 2

 ≥

(a + b + c)2 3(ab + bc + ca) 3 ≥ = 2(ab + bc + ca) 2(ab + bc + ca) 2

or, a b c √ +√ +√ ≥ c+a b+c a+b

r

3 2

Third Solution (Redwane Khyaoui): √

a b c 1 +√ +√ ≥ (a + b + c) 3 c + a b+c a+b

9

 √

1 1 1 +√ +√ c + a b+c a+b



f (x) =

√1 x

is a convex function, so jensen’s inequality gives: 



1 1 = LHS ≥ 3 · q 3 2 (a + b + c) 3

r

3 2

Problem 17 ‘—’ (keyree10): If a, b, c are REALS such that a2 + b2 + c2 = 1 Prove that √ a + b + c − 2abc ≤ 2

First Solution (Popa Alexandru): Use Cauchy-Schwartz: LHS = a(1 − 2bc) + (b + c) ≤

p (a2 + (b + c)2 )((1 − 2bc)2 + 1)

So it’ll be enough to prove that : (a2 + (b + c)2 )((1 − 2bc)2 + 1) ≤ 2 ⇔ (1 + 2bc)(1 − bc + 2b2 c2 ) ≤ 1 ⇔ 4b2 c2 ≤ 1 which is true because 1 ≥ b2 + c2 ≥ 2bc done

Problem 18 ‘—’ (Popa Alexandru): Let x, y, z > 0 such that xyz = 1 . Show that: x2 + y 2 + z 2 + x + y + z ≥ 2(xy + yz + zx)

First Solution (great math): To solve the problem of alex, we need Schur and Cauchy inequality as demonstrated as follow p √ 3 a + b + c ≥ 3 abc = 3 (abc)2 ≥

9abc ≥ 2(ab + bc + ca) − (a2 + b2 + c2 ) a+b+c

Note that we possess the another form of Schur such as (a2 + b2 + c2 )(a + b + c) + 9abc ≥ 2(ab + bc + ca)(a + b + c) Therefore, needless to say, we complete our proof here.

Problem 19 ‘Hoang Quoc Viet’ (Hoang Quoc Viet): Let a, b, c be positive reals satisfying a2 + b2 + c2 = 3. Prove that (abc)2 (a3 + b3 + c3 ) ≤ 3

10

First Solution (Hoang Quoc Viet): Let A = (abc)2 (a3 + b3 + c3 ) Therefore, we only need to maximize the following expression A3 = (abc)6 (a3 + b3 + c3 )3 Using Cauchy inequality as follows, we get
3





1 A = 6 3a2 b 3a2 c 3b2 a 3b2 c 3c2 a 3c2 b a3 + b3 + c3 ≤ 3 3



3(a2 + b2 + c2 )(a + b + c) 9

9

It is fairly straightforward that a+b+c≤

p

3(a2 + b2 + c2 ) = 3

Therefore, A3 ≤ 33 which leads to A ≤ 3 as desired. The equality case happens ⇐⇒ a = b = c = 1 Second Solution (FantasyLover): For the sake of convenience, let us introduce the new unknowns u, v, w as follows: u=a+b+c v = ab + bc + ca w = abc 2

3

3

3



2

 9 − u2 . 2

Now note that u − 2v = 3 and a + b + c = u(u − 3v) = u   9 − u2 2 We are to prove that w u · + 3w ≤ 3. 2 √ a+b+c u3 3 By AM-GM, we have abc ≤ =⇒ w ≤ 3 . 3 3 2 u9 7 9−u + 2 ≤ 37 . Hence, it suffices to prove that u · 2 3 r a+b+c a2 + b2 + c2 However, by QM-AM we have ≥ =⇒ u ≤ 3 3 3 6 2 7u (9 − u ) differentiating, u achieves its maximum when = 0. 2 Since a, b, c are positive, u cannot be 0, and the only possible value for u is 3. Since u ≤ 3, the above inequality is true.

Problem 20 ‘Murray Klamkin, IMO 1983’ (Hassan Al-Sibyani): Let a, b, c be the lengths of the sides of a triangle. Prove that: a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ 0

11

First Solution (Popa Alexandru): Use Ravi substitution a = x + y , b = y + z , c = z + x then the inequality becomes : y2 z2 x2 + + ≥x+y+z , y z x true by Cauchy-Schwartz. Second Solution (geniusbliss): we know from triangle inequality that b ≥ (a − c) and c ≥ (b − a) and a ≥ (c − b) therefore, a2 b(a − b) + b2 c(b − c) + c2 a(c − a) ≥ a2 (a − c)(a − b) + b2 (b − a)(b − c) + c2 (c − b)(c − a) ≥ 0 and the last one is schur’s inequality for r = 2 so proved with the equality holding when a = b = c or for and equilateral triangle

 Problem 21 ‘Popa Alexandru’ (Popa Alexandru): Let x, y, z ∈ 1≥

√ 3

xyz +

 1 2 , . Show that : 3 3

2 3(x + y + z)

First Solution (Endrit Fejzullahu): √ By Am-Gm x + y + z ≥ 3 3 xyz ,then 2 2√ √ √ + 3 xyz ≤ 3 xyz + 3 xyz 3(x + y + z)   9 2 1 2 1 √ Since x, y, z ∈ , ,then ≤ 3 xyz ≤ 3 3 3 3    2 1 2 √ 2 3 Let a = xyz ,then a + ≤ 1 ⇐⇒ 9a − 9a + 2 ≤ 0 ⇐⇒ 9 a − a− ≤ 0 ,we’re done since 9a 3 3 1 2 ≤a≤ 3 3

Problem 22 ‘Endrit Fejzullahu’ (Endrit Fejzullahu): Let a, b, c be side lengths of a triangle,and β is the angle between a and c.Prove that √ 2 3c sin β − a b2 + c2 > a2 b+c

First Solution (Endrit Fejzullahu): According to the Weitzenbock’s inequality we have √ ac sin β a2 + b2 + c2 ≥ 4 3S and S = 2 Then √ 2 2 2 b + c ≥ 2 √3ac sin β − a ,dividing by a2 , we have b2 + c2 2 3c sin β − a ≥ 2 a a √ √ b2 + c2 2 3c sin β − a 2 3c sin β − a Since a < b + c =⇒ ≥ > a2 a b+c

12

Problem 23 ‘Dinu Serbanescu, Junior TST 2002, Romania’ (Hassan Al-Sibyani): If a, b, c ∈ (0, 1) Prove that: p √ abc + (1 − a)(1 − b)(1 − c) < 1

First Solution (FantasyLover):

 π . Since a, b, c ∈ (0, 1), let us have a = sin2 A, b = sin2 B, c = sin2 C where A, B, C ∈ 0, 2 Then, we are to prove that sin A sin B sin C + cos A cos B cos C < 1. Now noting that sin C, cos C < 1, we have sin A sin B sin C + cos A cos B cos C < sin A sin B + cos A cos B = cos(A − B) ≤ 1.  Second Solution (Popa Alexandru): Cauchy-Schwartz and AM-GM works fine : p p √ √ √ √ abc + (1 − a) (1 − b) (1 − c) = a bc + 1 − a (1 − b) (1 − c) ≤ p p p a + (1 − a) bc + (1 − b) (1 − c) = bc + (1 − b) (1 − c) < 1

Third Solution (Redwane Khyaoui): √ abc +

p

(1 − a)(1 − b)(1 − c) ≤

a + bc 1 − a + (1 − b)(1 − c) + 2 2

a + bc 1 − a + (1 − b)(1 − c) + ≤1 So we only need to prove that 2 2 1 1 ⇔ + ≥ 2 which is true since a, b ∈ [0, 1] b c

Problem 24 ‘—’ (FantasyLover): For all positive real numbers a, b, c, prove the following: 1 1 a+1

+

1 b+1

+

1 c+1

−

1 1 a

+

1 b

+

1 c

≥

1 3

First Solution (Popa Alexandru): Using p, q, r substitution (p = a + b + c , q = ab + bc + ca , r = abc) the inequality becomes : 9 + 3r 3(p + q + r + 1) ≥ ⇔ pq + 2q 2 ≥ 6pr + 9r 2p + q + r q which is true because is well-known that pq ≥ 9r and q 2 ≥ 3pr Second Solution (Endrit Fejzullahu): After expanding the inequality is equivalent with :    1 1 1 1 1 1 1 1 1 1 + + ≥ + + + + a(a + 1) b(b + 1) c(c + 1) 3 a b c a+1 b+1 c+1 This is true by Chebyshev’s inequality , so we’re done .

13

Problem 25 ‘Mihai Opincariu’ (Popa Alexandru): Let a, b, c > 0 such that abc = 1 . Prove that : a2

ab bc ca √ + 2 √ + √ ≤1 2 2 2 + b + c b + c + a c + a2 + b

First Solution (FantasyLover): √ √ We have a2 + b2 + c ≥ 2ab + c = X Hence, it suffices to prove that 2

2 c

+

√

c.

1 a

X 1 √ ≤ 1. √ = 2 + a a + a cyc a cyc Reducing to a common denominator, we prove that √ √ √ √ √ √ X 1 4(a a + b b + c c) + ab ab + bc bc + ca ca + 12 √ = √ √ √ ≤1 √ √ √ 2+a a 4(a a + b b + c c) + 2(ab ab + bc bc + ca ca) + 9 cyc . √ √ √ Rearranging, it remains to prove that ab ab + bc bc + ca ca ≥ 3. p √ √ √ √ 3 Applying AM-GM, we have ab ab + bc bc + ca ca ≥ 3 a2 b2 c2 a2 b2 c2 = 3, and we are done.  Second Solution (Popa Alexandru): LHS ≤

X cyc

X X 1 ab 1 √ = √ = 2+ 2ab + c 2+c c cyc cyc

x y

≤ RHS

Problem 26 ‘Korea 2006 First Examination’ (FantasyLover): x, y, z are real numbers satisfying the condition 3x + 2y + z = 1. Find the maximum value of 1 1 1 + + 1 + |x| 1 + |y| 1 + |z| . First Solution (dgreenb801): We can assume x,y, and z are all positive, because if one was negative we could just make it positive, which would allow us to lessen the other two variables, making the whole sum larger. Let 3x = a, 2y = b, z = c, then a + b + c = 1 and we have to maximize 2 1 3 + + a+3 b+2 c+1 Note that     3 3 1 a2 c + ac2 + 6ac + 6c +1 − + = ≥0 a+c+3 a+3 c+1 (a + 3)(c + 1)(a + c + 3) So for fixed a + c, the sum is maximized when c = 0. We can apply the same reasoning to show the sum is maximized when b = 0. 11 So the maximum occurs when a = 1, b = 0, c = 0, and the sum is . 4

14

Problem 27 ‘Balkan Mathematical Olympiad 2006’ (dgreenb801): 1 1 1 3 + + ≥ a(1 + b) b(1 + c) c(1 + a) 1 + abc for all positive reals.

First Solution (Popa Alexandru): AM-GM works : (1 + abc) LHS + 3 =

X 1 + abc + a + ab cyc

a + ab

=

X 1+a X b(c + 1) √ 3 3 + ≥ √ + 3 abc ≥ 6 3 ab + a cyc b + 1 abc cyc

Problem 28 ‘Junior TST 2007, Romania’ (Popa Alexandru): Let a, b, c > 0 such that ab + bc + ca = 3. Show that : 1 1 1 1 + + ≤ 1 + a2 (b + c) 1 + b2 (c + a) 1 + c2 (a + b) abc

First Solution (Endrit Fejzullahu): Since ab + bc + ca = 3 =⇒ abc ≤ 1 Then 1 + 3a − abc ≥ 3a Then   X 1 1 1 1 1 1 ab + bc + ca ≤ + + = = 1 + 3a − abc 3 a b c 3abc abc

Problem 29 ‘Lithuania 1987’ (Endrit Fejzullahu): Let a, b, c be positive real numbers .Prove that a3 b3 c3 a+b+c + + ≥ 2 2 2 2 2 2 a + ab + b b + bc + c c + ca + a 3

First Solution (dgreenb801): By Cauchy, X X a3 a4 (a2 + b2 + c2 )2 = ≥ a2 + ab + b2 a3 + a2 b + ab2 a3 + b3 + c3 + a2 b + ab2 + b2 c + bc2 + c2 a + ca2 a+b+c This is ≥ if 3 (a4 + b4 + c4 ) + 2(a2 b2 + b2 c2 + c2 a2 ) ≥ (a3 b + a3 c + b3 a + b3 c + c3 a + c3 b) + (a2 bc + ab2 c + abc2 ) This is equivalent to (a2 + b2 + c2 )(a2 + b2 + c2 − ab − bc − ca) ≥ 0 Which is true as a2 + b2 + c2 − ab − bc − ca ≥ 0 ⇐⇒ (a − b)2 + (b − c)2 + (c − a)2 ≥ 0 Second Solution (Popa Alexandru): Since : a2

a3 − b3 =a−b + ab + b2

15

and similars we get : X cyc

X b3 1 X a3 + b3 a3 = = . a2 + ab + b2 a2 + ab + b2 2 cyc a2 + ab + b2 cyc

Now it remains to prove : 1 a+b a3 + b3 ≥ · 2 2 a + ab + b2 6 which is trivial .

√

Problem 30 ‘—’ (dgreenb801): Given ab + bc + ca = 1 Show that:

3a2 +b2 ab

√

+

3b2 +c2 bc

√

+

3c2 +a2 ca

√ ≥6 3

First Solution (Hoang Quoc Viet): Let’s make use of Cauchy-Schwarz as demonstrated as follows p (3a2 + b2 )(3 + 1) 3a + b ≥ 2ab 2ab Thus, we have the following estimations √ X cyc

3a2 + b2 ≥2 ab

Finally, we got to prove that



1 1 1 + + a b c



X1 √ ≥3 3 a cyc

However, from the given condition, we derive 1 abc ≤ √ 3 3 and

r X1 √ 1 3 ≥3 ≥3 3 a abc cyc

Problem 31 ‘Komal Magazine’ (Hoang Quoc Viet): Let a, b, c be real numbers. Prove that the following inequality holds (a2 + 2)(b2 + 2)(c2 + 2) ≥ 3(a + b + c)2

First Solution (Popa Alexandru): Cauchy-Schwartz gives : 1 3 (a2 + 2)(b2 + 2) = (a2 + 1)(1 + b2 ) + a2 + b2 + 3 ≥ (a + b)2 + (a + b)2 + 3 = ((a + b)2 + 2) 2 2 And Cauchy-Schwartz again (a2 + 2)(b2 + 2)(c2 + 2) ≥

√ 3 3 √ ((a + b)2 + 2)(2 + c2 ) ≥ ( 2(a + b) + 2c)2 = RHS 2 2

16

Problem 32 ‘mateforum.ro’ (Popa Alexandru): Let a, b, c ≥ 0 and a + b + c = 1 . Prove that : √

b2

b c 1 a +√ +√ ≥√ 2 2 1 + 3abc + 3c c + 3a a + 3b

First Solution (Endrit Fejzullahu): Using Holder’s inequality !2 X X a √ · a(b2 + 3c) ≥ (a + b + c)3 = 1 2 + 3c b cyc cyc It is enoughX to prove that 1 + 3abc ≥ a(b2 + 3c) cyc

Homogenise (a + b + c)3 = 1, X X Also after Homogenising a(b2 + 3c) = a2 b + b2 c + c2 a + 9abc + 3 a2 b 3

3

3

cyc 3

(a + b + c) = a + b + c + 6abc + 3

sym

X

2

a b

sym

It is enough to prove that a3 + b3 + c3 ≥ a2 b + b2 c + c2 a By AM-GM a3 + a3 + b3 ≥ 3a2 b b3 + b3 + c3 ≥ 3b2 c c3 + c3 + a3 ≥ 3c2 a Then a3 + b3 + c3 ≥ a2 b + b2 c + c2 a ,done Second Solution (manlio): by holder. P P (LHS)2 ( a(b2 + ac) ≥ ( a)3 so it suffices to prove 1 1 P ≥ a(b2 + 3c) 1 + 3abc that P 3is P 2 a ≥ ab true by AM-GM Third Solution (Apartim De): f (t) = √1t ; f 0 (t) < 0; f 00 (t) > 0 Using Jensen with weights a, b, c, we have af (b2 + 3c) + bf (c2 + 3a) + cf (a2 + 3b) ≥ f (ab2 + bc2 + ca2 + 3ab + 3bc + 3ca) Now, p By Holder, (a3 + b3 + c3 ) = 3 (a3 + b3 + c3 )(b3 + c3 + a3 )(b3 + c3 + a3 ) ≥ ab2 + bc2 + ca2 Again, P 2 3(a + b)(b + c)(c + a) + 3abc = 9abc + 3 a b = 3(a + b + c)(ab + bc + ca) = 3(ab + bc + ca) sym

∴ 1 + 3abc = (a + b + c)3 + 3abc > ab2 + bc2 + ca2 + 3ab + 3bc + 3ca ∴ f (ab2 + bc2 + ca2 + 3ab + 3bc + 3ca) > f (1 + 3abc) QED

17

Problem 33 ‘Apartim De’ (Apartim De): If a, b, c, d be positive reals then prove that: s a2 + b2 b2 + c 2 c2 + d2 54a + + ≥ 3 2 2 2 ab + b bc + c cd + d (a + d)

First Solution (Agr 94 Math): ( a )2 + 1 Write the LHS as ba + two other similar termsfeel lazy to write them down. This is a beautiful appli(b) + 1  2 t−1 t2 + 1 ≥0. is convex since cation of Jensen’s as the function for positve real t such that f (t) = t+1 t+1  P a d 2 ( b−a +1 3 Thus, we get that LHS ≥ P a d b − a +3 a b c I would like to write + + = K for my convenience with latexing. c d
b K 2 9 ) + 1 K +K 6 = ≥ so we have 3 3 3 K +3 1+ K 1+ K 9 This is from K + ≥ 6 by AM GM. K   13 Xa d d by AM GM. Now K = − ≥3 b a a  1 1 1 6 6a 3 3a 3 54a 3 Thus, we have ≥ ≥ = 1 1 1 3 a+d a+d 3 1+ K a3 + d3 2

Problem 34 ‘—’ (Raghav Grover): If a and b are non negative real numbers such that a ≥ b. Prove that 1 a+ ≥3 b(a − b)

First Solution (Dimitris X): If a ≥ 3 the problem is obviously true. Now for a < 3 we have : a2 b − ab2 − 3ab + 3b2 + 1 ≥ 0 ⇐⇒ ba2 − (b2 + 3b)a + 3b2 + 1 ≥ 0 It suffices to prove that D ≤ 0 ⇐⇒ b4 + 6b3 + 9b2 − 12b3 − 4b ⇐⇒ b(b − 1)2 (b − 4) ≤ 0 which is true.

Second Solution (dgreenb801): 1 1 a+ = b + (a − b) + ≥ 3 by AM-GM b(a − b) b(a − b) Third Solution (geniusbliss): p since a ≥ b we have a2 ≥ b(a − b) square this and substitute in this denominator we get, a a 4 + + ≥ 3 by AM-GM so done. 2 2 a2

18

Problem 35 ‘Vasile Cirtoaje’ (Dimitris X): √ √ √ (a2 − bc) b + c + (b2 − ca) c + a + (c2 − ab) a + b ≥ 0

2 First Solution (Popa Alexandru): Denote a+b 2 = x , ... , then the inequality becomes : X X xy(x3 + y 3 ) ≥ x2 y 2 (x + y) cyc

cyc

which is equivalent with : X

xy(x + y)(x − y)2 ≥ 0

cyc

Problem 36 ‘Cezar Lupu’ (Popa Alexandru): Let a, b, c > 0 such that abc ≥

p

1 a

+

1 b

+

1 c

=

√

abc. Prove that:

3(a + b + c)

First Solution√(Endrit Fejzullahu): √ 1 1 1 abc ⇐⇒ ab + bc + ca = abc abc a + b + c = By Am-Gm (ab + bc + ca)2 ≥ 3abc(a + b√+ c) Since ab + bc + ca = abc abc, wehave p (abc) 3 ≥ 3abc(a + b + c) =⇒ abc ≥ 3(a + b + c)

Problem 37 ‘Pham Kim Hung’ (Endrit Fejzullahu): Let a, b, c, d be positive real numbers satisfying a + b + c + d = 4.Prove that 1 1 1 1 1 + + + ≤ 2 2 2 2 11 + a 11 + b 11 + c 11 + d 3

First Solution (Apartim De): 6(x2 − 11 1 00 3 ) f (x) = ⇔ f (x) = 2 2 3 11 + x (11 + x )! r ! r 11 11 11 (x2 − ) = x − x+ 3 3 3 r r ! 11 11 If x ∈ − , , f 00 (x) < 0 3 3

19

r Thus within the interval

−

11 , 3

r

11 3

! , the quadratic polynomial is negative r

00

thereby making f (x) < 0, and thus f (x) is concave within r Let a ≤ b ≤ c ≤ d . If all of a, b, c, d ∈

0,

11 3

Then by Jensen, f (a) + f (b) + f (c) + f (d) ≤ 4f −2x 3

(11 + x2 )

r

11 3

! .

! , 

f 0 (x) =

−

11 , 3

a+b+c+d 4

 = 4f (1) =

4 1 = 12 3

< 0 (for all positive x) r

At most 2 of a, b, c, d(namely c& d) can be greater than

11 3

In that case,  f (a) + f (b) + f (c) + f (d) < f (a − 1) + f (b − 1) + f (c − 3) + f (d − 3) < 4f

a+b+c+d−8 4

 = 4f (−1) =

QED

Problem 38 ‘Crux Mathematicorum’ (Apartim De): Let R, r, s be the circumradius, inradius, and semiperimeter, respectively, of an acute-angled triangle. Prove or disprove that s2 ≥ 2R2 + 8Rr + 3r2 . When does equality occur?

First Solution (Virgil Nicula):

2




a + b2 + c2 = 2 · p2 − r2 − 4Rr (1)




2 2 2

4S = b + c − a · tan A (2)

. ThereProof. I”ll use the remarkable linear-angled identities

sin 2A + sin 2B + sin 2C = 2S2 (3)

R



cos A + cos B + cos C = 1 + r (4) R fore,
2 P 2 X X X cos A X cos2 A C.B.S.
1 + Rr ( cos A) (3)∧(4) 2 2 2 2 (2) a = b +c −a = 4S· = 8S· ≥ 8S· P = 8S· = 2S sin A sin 2A sin 2A R2
(1) = 4(R+r)2 =⇒ a2 + b2 + c2 ≥ 4(R + r)2 =⇒ 2· p2 − r2 − 4Rr ≥ 4(R+r)2 =⇒ p2 ≥ 2R2 + 8Rr + 3r2 .

Problem 39 ‘Russia 1978’ (Endrit Fejzullahu): Let 0 < a < b and xi ∈ [a, b].Prove that   1 1 1 n2 (a + b)2 (x1 + x2 + ... + xn ) + + ... + ≤ x1 x2 xn 4ab

20

4 1 = 12 3

First Solution (Popa Alexandru): We will prove that if a1 , a2 , . . . , an ∈ [a, b](0 < a < b) then   1 1 1 (a + b)2 2 (a1 + a2 + · · · + an ) + + ··· + ≤ n a1 a2 an 4ab      1 1 1 a1 a2 an  c c c P = (a1 + a2 + · · · + an ) + + ··· + = + + ··· + + + ··· + ≤ a1 a2 an c c c a1 a2 an  2 c a2 c an c 1 a1 + + + + ··· + + ≤ 4 c a1 c a2 c an √ c t Function f (t) = + have its maximum on [a, b] in a or b. We will choose c such that f (a) = f (b), c = ab. c r t r a b Then f (t) ≤ + . Then b a P ≤n

2

r

a + b

r

b b a

!2 ·

1 (a + b)2 = n2 4 4ab

Problem 40 ‘—’ (keyree10): a, b, c are non-negative reals. Prove that 81abc · (a2 + b2 + c2 ) ≤ (a + b + c)5

First Solution (Popa Alexandru): 81abc(a2 + b2 + c2 ) ≤ 27

(ab + bc + ca)2 2 (a + b2 + c2 ) ≤ (a + b + c)5 a+b+c

By p,q,r the last one is equivalent with : p6 − 27q 2 p2 + 54q 3 ≥ 0 ⇔ (p2 − 3q)2 ≥ 0

Problem 41 ‘mateforum.ro’ (Popa Alexandru): Let a, b, c > 0 such that a3 + b3 + 3c = 5 . Prove that : r r r a+b b+c c+a 1 1 1 + + ≤ + + 2c 2a 2b a b c

First Solution (Sayan Mukherjee): a3 + 1 + 1 +b3 + 1 + 1+ 3c = 9 ≥ 3a + 3b + 3c =⇒ a + b + c ≤ 3(AM-GM) a+b+c Also abc ≤ = 1(AM-GM) 3

21

Applying CS on the LHS; " #2 r X  X a+b 1 ≤ (a + b + c) 2c a X1 X It is left to prove that ≥ a a X1 But this =⇒ ≥3 a Which is perfextly true, as from AM-GM on the LHS; r X1 1 3 ≥3 ≥ 3 [∵ abc ≤ 1] a abc

Problem 42 ‘—’ (Sayan Mukherjee): Let a, b, c > 0 PT: If a, b, c satisfy ways have: X

X

1 1 = Then we ala2 + 1 2

1 1 ≤ a3 + 2 3

First Solution (Endrit Fejzullahu): a3 + a3 + 1 ≥ 3a2 =⇒ a3 + a3 + 1 + 3 ≥ 3a2 + 3 =⇒

1 1 1 X 1 ≥ =⇒ ≥ 3(a2 + 1) 2(a3 + 2) 3 a3 + 2

Problem 43 ‘Russia 2002’ (Endrit Fejzullahu): Let a, b, c be positive real numbers with sum 3.Prove that √ √ √ a + b + c ≥ ab + bc + ca

First Solution (Sayan Mukherjee): 2(ab + bc + ca) = 9 − (a2 + b2 + c2 ) as a + X b + c = 9. Hence the inequality √ a ≥ 9 − (a2 + b2 + c2 ) =⇒ 2 X √ √ =⇒ (a2 + a + a) ≥ 9 Perfectly √ true √ from AGM as : a2 + a + a ≥ 3a

Problem 44 ‘India 2002’ (Raghav Grover): For any natural number n prove that 1 1 1 1 1 1 ≤ 2 + 2 + .... + 2 ≤ + 2 n +1 n +2 n +n 2 2n

22

First Solution (Sayan Mukherjee): Pn n n X X k 3(n2 + n) k 1 k2 k=1 P P = ≥ = > n n 2 2 2 2 2 2 n +k n k+k 2(3n + 2n + 1) 2 n k=1 k + k=1 k k=1

k=1

As it is equivalent to :3n2 + 3n > 3n2 + 2n + 1 =⇒ n > 1 For the 2nd part; X X 1 X k X 1 n2 = 1− 2 = n − n2 and then use AM-HM for So we get the desired 2 2 n +k n +k n +k n2 + k result.

Problem 45 ‘—’ (Sayan Mukherjee): For a, b, c > 0; a2 + b2 + c2 = 1 find Pmin if: P =

X a2 b2 cyc

c2

First Solution (Endrit Fejzullahu): ab bc ca Let x = , y = andz = c a b Then Obviously By Am-Gm x2 + y 2 + z 2 ≥ xy + yz + zx = a2 + b2 + c2 = 1 ,then Pmin = 1

Problem 46 ‘—’ (Endrit Fejzullahu): Let a, b, c be positive real numbers such that a + b + c = 3 .Prove that a2 b2 c2 + + ≥1 a + 2b2 b + 2c2 c + 2a2

First Solution (Popa Alexandru): We start with a nice use of AM-GM : a2 a2 + 2ab2 − 2ab2 a(a + 2b2 ) 2ab2 2√ 3 = = − ≥a− a2 b2 2 2 2 2 a + 2b a + 2b a + 2b a + 2b 3 Suming the similars we need to prove : √ 3

a 2 b2 +

√ 3

b2 c 2 +

√ 3

c2 a2 ≤ 3

By AM-GM : X√ 3 cyc

a2 b2

≤

X 2ab + 1 cyc

3

1 = 3

! 2

X

ab + 3

≤ 3 ⇔ ab + bc + ca ≤ 3 ⇔ 3(ab + bc + ca) ≤ (a + b + c)2

cyc

Problem 47 ‘mateforum.ro’ (Popa Alexandru): Let a, b, c > 0 such that a + b + c ≤ : a b c a b c + + ≤ + + b+c c+a a+b (b + c)2 (c + a)2 (a + b)2

23

3 2

. Prove that

First Solution: WLOG a ≥ b ≥ c Then By Chebyshev’s  inequality we have  1 1 1 1 + + RHS ≥ · LHS · 3 a+b b+c c+a It is enough to show that 1 1 1 + + ≥3 a+b b+c c+a By Cauchy Schwartz 1 1 1 9 3 + + ≥ ≥ 3 ⇐⇒ a + b + c ≤ , done a+b b+c c+a 2(a + b + c) 2

Problem 48 ‘USAMO 2003’ (Hassan Al-Sibyani): Let a, b, c be positive real numbers. Prove that: (2b + a + c)2 (2c + a + b)2 (2a + b + c)2 + + ≤8 2a2 + (b + c)2 2b2 + (a + c)2 2c2 + (a + b)2

First Solution (Popa Alexandru): Suppose a + b + c = 3 The inequality is equivalent with : (a + 3)2 (b + 3)2 (c + 3)2 + 2 + 2 ≤8 2 2 + (3 − a) 2b + (3 − b) 2c + (3 − c)2

2a2 For this we prove :

(a + 3)2 4 4 ≤ a + ⇔ 3(4a + 3)(a − 1)2 ≥ 0 + (3 − a)2 3 3

2a2 done.

Problem 49 ‘Marius Mainea’ (Popa Alexandru): Let x, y, z > 0 such that x + y + z = xyz. Prove that : x+y y+z z+x 27 + + ≥ 1 + z2 1 + x2 1 + y2 2xyz

First Solution (socrates): Using Cauchy-Schwarz and AM-GM inequality we have: x+y y+z z+x (x + y)2 (y + z)2 (z + x)2 ((x + y) + (y + z) + (z + x))2 P + + = + + ≥ = 2 2 2 2 2 2 1+z 1+x 1+y x + y + (x + y)z y + z + (y + z)x z + x + (z + x)y 2(x + y + z) + x2 (y + z) 4(x + y + z)2 4(x + y + z)2 27 1 27 P 2 = Q ≥ 4(x + y + z)2 · = 3 2xyz + x (y + z) (x + y) 8 (x + y + z) 2xyz

24

Problem 50 ‘—’ (socrates): Let a, b, c > 0 such that ab + bc + ca = 1 . Prove that : p p p abc(a + a2 + 1)(b + b2 + 1)(c + c2 + 1) ≤ 1

p √ √ 2+1= 2 + ab + bc + ca = First Solution (dgreenb801): Note that a a (a + b)(a + c) p 2 Also, by Cauchy, (a + (a + b)(a + c)) ≤ (a + (a + b))(a + (a + c)) = (2a + b)(2a + c) So after squaring both sides of the inequality, we have to show 1 = (ab + bc + ca)6 ≥ a2 b2 c2 (2a + b)(2a + c)(2b + a)(2b + c)(2c + a)(2c + b) = (2ac + bc)(2ab + bc)(2bc + ac)(2ab + ac)(2cb + ab)(2ca + ab) , which is true by AM-GM.

Problem 51 ‘Asian Pacific Mathematics Olympiad’ (dgreenb801): Let a, b, c > 0 such that abc = 8. Prove that: a2 b2 c2 4 p +p +p ≥ 3 (1 + a3 )(1 + b3 ) (1 + b3 )(1 + c3 ) (1 + c3 )(1 + a3 ) First Solution (Popa Alexandru): By AM-GM : p Then we have : X cyc

a3 + 1 ≤

a2 + 2 . 2

X a2 a2 p ≥4 (a2 + 2) (b2 + 2) (a3 + 1) (b3 + 1) cyc

So we need to prove X cyc

a2 1 ≥ . 2 2 (a + 2) (b + 2) 3

which is equivalent with : a2 b2 + b2 c2 + c2 a2 + 2(a2 + b2 + c2 ) ≥ 72. , true by AM-GM

Problem 52 ‘Lucian Petrescu’ (Popa Alexandru): Prove that in any acute-angled triangle ABC we have : a+b b+c c+a + + ≥ 4(a + b + c) cos C cos A cos B

First Solution (socrates): The inequality can be rewritten as ab(a + b) bc(b + c) ca(c + a) + 2 + 2 ≥ 2(a + b + c) 2 2 2 2 +b −c b +c −a c + a2 − b2

a2

25

or X

a(

cyclic

a2

c2 b2 + 2 ) ≥ 2(a + b + c) 2 2 +b −c c + a2 − b2

Applying Cauchy Schwarz inequality we get X

a(

cyclic

a2

X b2 c2 (b + c)2 + 2 )≥ a 2 2 2 2 2 2 +b −c c +a −b (a + b − c2 ) + (c2 + a2 − b2 )

or X

a(

cyclic

a2

X (b + c)2 c2 b2 + 2 )≥ 2 2 2 2 +b −c c +a −b 2a

So, it is enough to prove that X (b + c)2 2a

≥ 2(a + b + c)

which is just CS as above. Second Solution (Mateescu Constantin): From the law of cosinus we have a = b cos C + c cos Band c = b cos A + a cos B . a+c b(cos A + cos C) =⇒ =a+c+ = a + c + 2R tan B(cos A + cos C) . Then: B cosX B X acos +c = 2(a + b + c) + 2R tan B · (cos A + cos C) and the inequality becomes: cos B X X X 2R tan B(cos A + cos C) ≥ 2 a = 4R sin A X X X X ⇐⇒ tan B cos A + tan B cos C ≥ 2 sin A = 2 tan A cos A Wlog assume that A ≤ B ≤ C . Then cos A ≥ cos B ≥ cos C and tan A ≤ tan B ≤ tan C, so X X P P tan A cos A ≤ tan B cos A and tan A cos A ≤ tan B cos C, according to rearrangement inequality . Adding up these 2 inequalities yields the conclusion. Third Solution (Popa Alexandru): LHS ≥

(2(a + b + c))2 = 4(a + b + c) (a + b) cos C + (b + c) cos A + (c + a) cos B

Problem 53 ‘—’ (socrates): Given x1 , x2 , .., xn > 0 such that

n X i=1

n X xi + n i=1

1 + x2i

≤ n2

First Solution (Hoang Quoc Viet): Without loss of generality, we may assume that x1 ≥ x2 ≥ · · · ≥ xn

26

xi = 1, prove that

Therefore, we have nx1 − 1 ≥ nx2 − 1 ≥ · · · ≥ nxn − 1 and

x2 xn x1 ≥ 2 ≥ ··· ≥ 2 x21 + 1 x2 + 1 xn + 1

Hence, by Chebyshev inequality, we get n X (nxi − 1)xi i=1

x2i + 1

# n ! " ! n X X xi 1 ≥ n xi − n =0 n x2 + 1 i=1 i=1 i

Thus, we get the desired result, which is n X xi + n i=1

1 + x2i

≤ n2

Problem 54 ‘Hoang Quoc Viet’ (Hoang Quoc Viet): Let a, b, c be positive reals satisfying a2 + b2 + c2 = 3. Prove that a3 b3 c3 + 2 + 2 ≥1 2 2 2 2b + c 2c + a 2a + b2

First Solution (Hoang Quoc Viet): Using Cauchy Schwartz, we get X cyc

a3 (a2 + b2 + c2 )2 P P ≥ 2b2 + c2 2 cyc ab2 + cyc ac2

Hence, it suffices to prove that ab2 + bc2 + ca2 ≤ 3 and a2 b + b2 c + c2 a ≤ 3 However, applying Cauchy Schwartz again, we obtain p a(ab) + b(bc) + c(ca) ≤ (a2 + b2 + c2 ) ((ab)2 + (bc)2 + (ca)2 ) In addition to that, we have (ab)2 + (bc)2 + (ca)2 ≤

(a2 + b2 + c2 )2 3

Hence, we complete our proof.

Problem 55 ‘Iran 1998’ (saif): Let x, y, z > 1 such that √

x+y+z ≥

√

x−1+

27

1 x

√

+

1 y

+

x−1+

1 z

= 2 prove that:

√

z−1

First Solution (beautifulliar):√ √ √ √ Note that you can substitute x − 1 = a, y − 1 = b, z − 1 = c then you need to prove that a2 + b2 + c2 + 3 ≥ 1 1 1 a + b + c while you have 2 + + = 2 or equivalently a2 b2 + b2 c2 + c2 a2 + 2a2 b2 c2 = 1. next, a + 1 b2 + 1 c2 + 1 substitute ab = cos x, bc = cos y, ca = cos z where x, y, z are angles of triangle. since you need to prove that √ 3 a2 + b2 + c2 + 3 ≥ a+b+c then you only need to prove that 3 ≥ 2(ab+bc+ca) or cos x+cos B +cos C ≤ 2 which is trivial. Second Solution (Sayan Mukherjee): Xx−1 1 1 1 + + = 2 =⇒ =1 x y z x 2 X x − 1 X √ x − 1 (From CS) Hence (x + y + z) ≥ X √x √ =⇒ x + y + z ≥ x−1

Problem 56 ‘IMO 1998’ (saif): Let a1 , a2 , ..., an > 0 such that a1 + a2 + ... + an < 1. prove that 1 a1 .a2 ...an (1 − a1 − a2 − ... − an ) ≤ n−1 (a1 + a2 + ... + an )(1 − a1 )(1 − a2 )...(1 − an ) n

First Solution (beautifulliar): 1 a1 a2 . . . an an+1 ≤ n+1 (it should be (1 − a1 )(1 − a2 ) . . . (1 − an ) n √ nn+1 right?) which follows from am-gm a1 + a2 + · · · + an ≥ n n a1 a2 . . . an , a2 + a3 + · · · + an+1 ≥ √ √ n n a2 a3 . . . an+1 , a1 + a3 + · · · + an+1 ≥ n n a1 a3 . . . an+1 , and so on... you will find it easy.

Let an+1 = 1 − a1 − a2 − · · · − an the we arrive at

Problem 57 ‘—’ (beautifulliar): Let n be a positive integer. If x1 , x2 , . . . , xn are real numbers such that x1 + x2 + · · · + xn = 0 and also y = max{x1 , x2 , . . . , xn } and also z = min{x1 , x2 , . . . , xn }, prove that x21 + x22 + · · · + x2n + nyz ≤ 0

First Solution (socrates): (xi − y)(xi − z) ≤ 0 ∀i = 1, 2, ..., n so

n X i=1

(x2i + yz) ≤

n X

(y + z)xi = 0 and the conclusion follows.

i=1

Problem 58 ‘Pham Kim Hung’ (Endrit Fejzullahu): Suppose that x, y, z are positive real numbers and x5 + y 5 + z 5 = 3.Prove that x4 y4 z4 + 3 + 3 ≥3 3 y z x

28

First Solution (Popa Alexandru): By a nice use of AM-GM we have :  4 p √ √ y4 z4 x 19 19 + 3 + 3 + 3(x5 + y 5 + z 5 )2 ≥ 19( x100 + 19 y 100 + z 100 ) 10 3 y z x So it remains to prove : √

3 + 19(

19

x100 +

p

19

y 100 +

√

19

z 100 ) ≥ 20(x5 + y 5 + z 5 )

which is true by AM-GM .

Problem 59 ‘China 2003’ (bokagadha): x, y, and z are positive real numbers such that x + y + z = xyz. Find the minimum value of: x7 (yz − 1) + y 7 (xz − 1) + z 7 (xy − 1)

First Solution (Popa Alexandru): By AM-GM and the condition you get

√ xyz ≥ 3 3

Also observe that the condition is equivalent with yz − 1 = So the

y+z x

p √ LHS = x6 (y + z) + y 6 (z + x) + z 6 (x + y) ≥ 6 6 x14 y 14 z 14 ≥ 216 3

Problem 60 ‘Austria 1990’ (Rofler): v s u r q u √ t 2 3 4 ... N < 3 − − − ∀N ∈ N≥2

r q p √ First Solution (Brut3Forc3): We prove the generalization m (m + 1) . . . N < m + 1, for m + 2. √ For m = N , this isrequivalent to N < N + 1, which is clearly true. We now induct from m = N q √ down. Assume that (k + 1) (k + 2) . . . N < k + 2. Multiplying by k and taking the square root gives r q p √ p k (k + 1) . . . N < k(k + 2) < k + 1, completing the induction.

Problem 61 ‘Tran Quoc Anh’ (Sayan Mukherjee): Given a, b, c ≥ 0 Prove that: r X a+b 3 ≥√ 2 + 4bc + c2 b a + b+c cyc

29

First Solution (Hoang Quoc Viet): Using Cauchy inequality, we get s s r X 2(a + b) X X a+b 8 ≥ ≥ 36 2 2 2 b + 4bc + c 3(b + c) 27(a + b)(b + c)(c + a) cyc cyc cyc However, we have (a + b)(b + c)(c + a) ≤

8(a + b + c)3 27

Hence, we complete our proof here.

Problem 62 ‘Popa Alexandru’ (Popa Alexandru): Let a, b, c > 0 such that a + b + c = 1 . Show that : a2 + ab b2 + bc c2 + ca 3 + + ≥ 1 − a2 1 − b2 1 − c2 4

First Solution (Endrit Fejzullahu): Let a + b = x , b + c = y , c + a = z ,given inequality becomes Xx y

+

X

x 9 ≥ x+y 2

Then By Cauchy Schwartz LHS =

Xx cyc

y

+

X cyc

P P ( x)4 8( x)4 9 x P P 2 ≥ P P 2 2 ≥ ≥ P x+y ( xy)( xy + x ) ( xy + ( x) ) 2

Second Solution (Hoang Quoc Viet): Without too many technical terms, we have ! Xx+y X x X rx + ≥ ≥3 4y x+y y cyc cyc cyc Therefore, it is sufficient to check that

Xx cyc

y

≥3

which is Cauchy inequality for 3 positive reals.

Problem 63 ‘India 2007’ (Sayan Mukherjee): For positive reals a, b, c. Prove that: (a + b + c)2 (ab + bc + ca)2 ≤ 3(a2 + ab + b2 )(b2 + bc + c2 )(c2 + ac + a2 )

First Solution (Endrit Fejzullahu): I’ve got an SOS representation of : RHS −LHS =

1 ((x+y +z)2 (x2 y 2 +y 2 z 2 +z 2 x2 −xyz(x+y +z))+(xy +yz +zx)(x2 +y 2 +z 2 −xy −yz −zx)) 2

30

So I may assume that the inequality is true for reals

Problem 64 ‘Popa Alexandru’ (Endrit Fejzullahu): Let a, b, c > 0 such that (a + b)(b + c)(c + a) = 1 . Show that : 3 2 16abc ≥a+b+c≥ ≥ 16abc 3 3 (a + b)(b + c)(c + a) 1 2 16 First Solution (Apartim De): By AM-GM, ≥ abc ⇔ ≥ abc ⇔ ≥ abc By 8 8 3 3 AM-GM, (a + b) + (b + c) + (c + a) ≥ 3 2 3 ⇔ (a + b + c) ≥ > 2 3 −−→ −−→ −−→ Lemma:We have for any positive reals x, y, z and vectors M A, M B, M C  −−→  −−→ −−→ 2 xM A + y M B + z M C ≥ 0
⇔ (x + y + z)(xM A2 + yM B 2 + zM C 2 ) ≥ xyAB 2 + yzBC 2 + zxCA2 Now taking x = y = z = 1 and M to be the circumcenter of the triangle with sides p, q, r such that pqr = 1 , and the area of the triangle=∆, we have by the above lemma, 2 9R2 ≥ p2 + q 2 + r2 ≥ 3 (pqr) 3 = 3 ⇔ 3R2 ≥ 1 ⇔ 16∆2 ≤ 3 ⇔ (p + q + r)(p + q − r)(q + r − p)(r + p − q) ≤ 3 Now plugging in the famous Ravi substitution i.e, p = (a + b); q = (b + c); r = (c + a) 3 ⇔ (a + b + c) ≤ 16abc Problem 65 ‘IMO 1988 shortlist’ (Apartim De): In the plane of the acute angled triangle ∆ABC, L is a line such that u, v, w are the lengths of the perpendiculars from A, B, C respectively to L. Prove that u2 tan A + v 2 tan B + w2 tan C ≥ 2∆ where ∆ is the area of the triangle. First Solution (Hassan Al-Sibyani): Consider a Cartesian system with the x-axis on the line BC and origin at the foot of the perpendicular from A to BC, so that A lies on the y-axis. Let A be (0, α), B(−β, 0), C(γ, 0), where α, β, γ > 0 (because ABC is acute-angled). Then α tan B = β α tan C = γ α(β + γ) tan A = − tan(B + C) = 2 α − βγ here tan A > 0,so α2 > βγ. Let L have equation x cos θ + y sin θ + p = 0 Then u2 tan A + v 2 tan B + w2 tan C α(β + γ) α α = 2 (α sin θ + p)2 + (−β cos θ + p)2 + (γ cos θ + p)2 α − βγ β γ α(β + γ) α(β + γ) 2 = α2 sin2 θ + 2αp sin θ + p2 ) 2 + α(β + γ) cos2 θ + p α − βγ βγ α(β + γ) = (α2 p2 + 2αpβγ sin θ + α2 βγ sin2 θ + βγ(α2 − βγ) cos2 θ) βγ(α2 − βγ)

31

α(β + γ) [(αp + βγ sin θ)2 + βγ(α2 − βγ)] ≥ α(β + γ) = 2∆ βγ(α2 − βγ) with equality when αp + βγ sin θ = 0, i.e., if and only if L passes through (0, βγ/α), which is the orthocenter of the triangle.

=

Problem 66 ‘—’ (Hassan Al-Sibyani): For positive real number a, b, c such that abc ≤ 1, Prove that: c a b + + ≥a+b+c b c a

First Solution (Endrit Fejzullahu): It is easy to prove that a b c a+b+c + + ≥ √ 3 b c a abc and since abc ≤ 1 then

a b

+

b c

+

c a

≥ a + b + c ,as desired .

Problem 67 ‘Endrit Fejzullahu’ (Endrit Fejzullahu): Let a, b, c, d be positive real numbers such that a + b + c + d = 4.Find the minimal value of : X cyc

a4 (b + 1)(c + 1)(d + 1)

First Solution (Sayan Mukherjee): From AM-GM,  X  X 4a a4 b+1 c+1 d+1 + + + ≥ P = (b + 1)(c + 1)(d + 1) 16 16 16 8 cyc cyc =⇒

X cyc

So Pmin =

 a4 3 3 a+b+c+d − (a + b + c + d) − 4 · ≥ (b + 1)(c + 1)(d + 1) 2 16 16   X a4 4 6 3 1 ∴ ≥ − =2− = (b + 1)(c + 1)(d + 1) 2 4 2 2 cyc

1 2

Problem 68 ‘—’ (Sayan Mukherjee): Prove that X cyc

x 1 ≤ ∀1 ≥ x, y, z ≥ 0 7 + z3 + y3 3

32

First Solution (Toang Huc Khein): X X X x x x 1 ≤ ≤ = because x3 + y 3 + z 3 + 6 ≥ 3(x + y + z) ⇔ 3 3 3 3 3 7 + y + z 6 + x + y + z 3(x + y + z) 3 X (x − 1)2 (x + 2) ≥ 0

Problem 69 ‘Marius Maine’ (Toang Huc Khein): Let x, y, z > 0 with x + y + z = 1 . Then : y 2 − zx z 2 − xy x2 − yz + + 2 ≤0 x2 + x y2 + y z +z

First Solution (Endrit Fejzullahu): x + yz x2 − yz =1− 2 x2 + x x +x Inequality is equivalent with :

X x + yz x2 + x

cyc

X x + yz cyc

x2

+x

=

X cyc

≥3

X yz 1 + x + 1 cyc x2 + x

By Cauchy-Schwarz inequality X cyc

1 9 ≥ x+1 4

And X cyc

(xy + yz + zx)2 (xy + yz + zx)2 3 yz ≥ = ≥ ⇐⇒ (xy + yz + zx)2 ≥ 3xyz x2 + x x2 yz + y 2 xz + z 2 xy + 3xyz 4xyz 4

This is true since x + y + z = 1 and (xy + yz + zx)2 ≥ 3xyz(x + y + z) Second Solution (Endrit Fejzullahu): Let us observe that by Cauchy-Schwartz we have : X x + yz cyc

x2 + x

=

X x(x + y + z) + yz cyc

x(x + 1)

=

X (x + y)(x + z) X = x(x + y + x + z) cyc cyc

x x+y

1 +

x x+z

≥

(1 + 1 + 1)2 9 = =3 3 3

Then we can conclude that : LHS =

X x2 + x cyc

x2

+x

−

X x + yz cyc

x2 + x

≤3−3=0

Problem 70 ‘Claudiu Mandrila’ (Endrit Fejzullahu): Let a, b, c > 0 such that abc = 1. Prove that : a10 b10 c10 a7 b7 c7 + + ≥ 7 + + b+c c+a a+b b + c7 c7 + a7 a7 + b7

33

First Solution (Sayan Mukherjee): Since abc = 1 so, we have: X X X a10 a7 = ≥ 3 3 b+c b c (b + c)

1 64 (b

a7 + c)6 · (b + c)

So we are only required to prove that: (b + c)7 ≤ 26 b7 + 26 c7 But, from Holder; 6 1 (b7 + c7 ) 7 (1 + 1) 7 ≥ b + c Hence we are done

Problem 71 ‘Hojoo Lee, Crux Mathematicorum’ (Sayan Mukherjee): Let a, b, c ∈ R+ ;Prove that: 2 3 9(a + b + c)2 ≥ 33 (a + b3 + c3 ) + 2 abc a + b2 + c2

First Solution (Endrit Fejzullahu): Inequality is equivalent with :

(a + b + c)(a2 + b2 + c2 ) − 9abc (a − b)2 + (b − c)2 + (c − a)2 ≥ 0

34