Brilliant Inequality Contest - Season 1

Welcome all to the first ever Brilliant Inequality Contest. Like the Brilliant Integration Contest, the aim of the Inequality Contest is to improve skills and techniques often used in Olympiad-style Inequality problems. But above all, the main reason is so we can have fun.

Anyone is allowed to participate, as long as they adhere to the following rules.

  1. I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

  2. A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

  3. Please make a substantial comment. Spam or unrelated comments will be deleted. (To help me out, try to delete your comments within an hour of when you posted them, if they are not the solutions.)

  4. Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 48 hours. Then, you must post the solution and you have the right to post a new problem.

  5. If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

  6. The scope of the problems is Olympiad-style inequalities.

  7. You are not allowed to post problems requiring calculus in the solutions (use of differentiation to prove a curve is concave or convex is allowed).

  8. Lagrange Multipliers are not allowed to be used in a solution.

  9. Inequalities allowed to be used are AM-GM, Muirhead, Power Mean and Weighted Power Mean, Cauchy-Schwarz, Holder, Rearrangement, Chebyshev, Schur, Jensen, Karamata, Reverse Rearrangement and Titu's lemma.

  10. Try to post the simplest solution possible. For example, if someone posted a solution using Holder, Titu's and Cauchy, when there is a solution using only AM-GM, the latter is preferred.

Format your proof as follows:

SOLUTION OF PROBLEM (insert problem number here)

[Post your solution here]


PROBLEM (insert problem number here)

[Post your problem here]

Remember to reshare this note so it goes to everyone out there. And above all else, have fun!!!


PROBLEM 1

Let xx, yy and zz be positive reals such that x+yzx+y \geq z, y+zxy+z \geq x and z+xyz+x \geq y. Find families of solutions for (x,y,z)(x, y, z) such that the following inequality is satisfied.

2x2(y+z)+2y2(z+x)+2z2(x+y)x3+y3+z3+9xyz2x^2 (y+z) + 2y^2 (z+x) + 2z^2 (x+y) \geq x^3 + y^3 + z^3 + 9xyz


P.S.: For those who want to discuss problem solutions, they can do so here.

Note by Sharky Kesa
3 years, 10 months ago

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1 vote

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Solution of Problem 1 :

We will show that the given inequality is true for all positive reals x,y,zx,y,z satisfying the conditions given in the question.

Case 1: When x,y,z are sides of triangle.

Let x=a+b,y=b+c,z=a+c\color{#3D99F6}{x=a+b,y=b+c,z=a+c} ; (a,b,c>0)\text{(a,b,c>0)} ,so our inequality becomes:

cyc2(a+b)2(a+b+2c)cyc(a+b)3+9(a+b)(b+c)(a+c)\color{#69047E}{\sum_{cyc}2(a+b)^2(a+b+2c) \geq \sum_{cyc} (a+b)^3 +9(a+b)(b+c)(a+c)}

    2cyc(a+b)3+4cycc(a+b)2cyc(a+b)3+9(a+b)(b+c)(a+c)\implies 2 \sum_{cyc}(a+b)^{3} + 4\sum_{cyc} c(a+b)^{2} \geq \sum_{cyc}(a+b)^3 +9(a+b)(b+c)(a+c)

    cyc(a+b)3+4cycc(a+b)29(a+b)(b+c)(a+c)\implies \sum_{cyc} (a+b)^{3} +4\sum_{cyc} c(a+b)^{2} \geq 9(a+b)(b+c)(a+c)

    2cyca3+3cyc(ab2+ba2)+4cyc(ab2+ba2)+24abc9(a+b)(b+c)(a+c)\implies 2 \sum_{cyc}a^{3}+ 3\sum_{cyc} (ab^{2} +ba^{2}) + 4\sum_{cyc} (ab^{2} +ba^{2}) + 24abc \geq 9(a+b)(b+c)(a+c)

    2cyca3+7(a+b)(b+c)(c+a)+10abc9(a+b)(b+c)(c+a)\implies 2 \sum_{cyc} a^{3} + 7(a+b)(b+c)(c+a) + 10abc \geq 9(a+b)(b+c)(c+a)

    2cyca3+10abc2(a+b)(b+c)(c+a)\implies 2 \sum_{cyc} a^{3}+ 10abc \geq 2(a+b)(b+c)(c+a)

    cyca3+5abc(a+b)(b+c)(c+a)\implies \sum_{cyc} a^{3}+ 5abc \geq (a+b)(b+c)(c+a)

    cyca3+3abca2b+b2a+a2c+c2a+b2c+c2b\color{#D61F06}{\implies \sum_{cyc} a^{3} + 3abc \geq a^{2} b + b^{2} a + a^{2} c + c^{2} a + b^{2} c + c^{2} b}

which is true by Schur's Inequality.

Equality occurs when a=b=ca=b=c , that is x=y=zx=y=z.

Hence proved.

Case 2: When x,y,z are not sides of a triangle.

In this case either x+y=zx+y=z or y+z=xy+z = x or z+x=yz+x=y.

Let's take x+y=z\color{#3D99F6}{x+y =z} .

Since the inequality is symmetric, WLOG we can assume that zxy\color{#20A900}{z \ge x \ge y}.

Now substituting x+y=zx+y = z in the inequality , we get 2x2(2y+x)+2y2(2x+y)+2(x+y)3x3+y3+(x+y)3+9xyz\color{#D61F06}{ 2x^{2}(2y+x) + 2y^{2}(2x+y) + 2(x+y)^{3} \geq x^{3} + y^{3} + (x+y)^{3} + 9xyz}

After rearranging the expression, we get,x3+y3x2y+y2x x^{3} + y^{3} \geq x^{2} y + y^{2}x

which is true by Rearrangement Inequality

Hence proved.

Harsh Shrivastava - 3 years, 10 months ago

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Can you write the equality cases for Case 2?

Sharky Kesa - 3 years, 10 months ago

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I don't think equality exists . . .

Harsh Shrivastava - 3 years, 10 months ago

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Problem 8

For positive reals a,b,ca,b,c, prove that (1a+b+1b+c)(1b+c+1c+a)(1c+a+1a+b)1abc\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\le \dfrac{1}{abc}

Daniel Liu - 3 years, 10 months ago

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Solution to Problem 8

I used a massive expansion, but I've triple-checked so its right (If you haven't figured it out, expansion is what I do with inequalities). We have to prove

(1a+b+1b+c)(1b+c+1c+a)(1c+a+1a+b)1abc\left (\dfrac {1}{a+b} + \dfrac{1}{b+c} \right ) \left (\dfrac {1}{b+c} + \dfrac{1}{c+a} \right ) \left (\dfrac {1}{c+a} + \dfrac{1}{a+b} \right ) \leq \dfrac {1}{abc}

(a+2b+c(a+b)(b+c))(b+2c+a(b+c)(c+a))(c+2a+b(c+a)(a+b))1abc\left (\dfrac {a+2b+c}{(a+b)(b+c)} \right ) \left (\dfrac {b+2c+a}{(b+c)(c+a)} \right ) \left (\dfrac {c+2a+b}{(c+a)(a+b)} \right ) \leq \dfrac {1}{abc}

Simplifying the denominators, we have

abc(a+2b+c)(b+2c+a)(c+2a+b)(a+b)2(b+c)2(c+a)2abc(a+2b+c)(b+2c+a)(c+2a+b) \leq (a+b)^2 (b+c)^2 (c+a)^2

Simplifying all the brackets, we get

2a4bc+2ab4c+2abc4+7a3b2c+7a3bc2+7a2b3c+7ab3c2+7a2bc3+7ab2c3+16a2b2c22a4bc+2ab4c+2abc4+6a3b2c+6a3bc2+6a2b3c+6ab3c2+6a2bc3+6ab2c3+10a2b2c2+a4b2+a4c2+b4a2+b4c2+c4a2+c4b2+2a3b3+2b3c3+2c3a3\begin{aligned} &2a^4bc + 2ab^4c + 2abc^4 + 7a^3b^2c + 7a^3bc^2 + 7a^2b^3c + 7ab^3c^2 + 7a^2bc^3 + 7ab^2c^3 + 16a^2b^2c^2 \leq\\ &2a^4bc + 2ab^4c + 2abc^4 + 6a^3b^2c + 6a^3bc^2 + 6a^2b^3c + 6ab^3c^2 + 6a^2bc^3 + 6ab^2c^3 + 10a^2b^2c^2 \\& \quad \quad \quad \quad \quad+a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2 + 2a^3b^3 + 2b^3c^3 + 2c^3a^3 \end{aligned}

Simplifying, we get

a3b2c+a3bc2+a2b3c+ab3c2+a2bc3+ab2c3+6a2b2c2a4b2+a4c2+b4a2+b4c2+c4a2+c4b2+2a3b3+2b3c3+2c3a3a^3b^2c + a^3bc^2 + a^2b^3c + ab^3c^2 + a^2bc^3 + ab^2c^3 + 6a^2b^2c^2 \leq a^4b^2 + a^4c^2 + b^4a^2 + b^4c^2 + c^4a^2 + c^4b^2+2a^3b^3+2b^3c^3+2c^3a^3

By AM - GM, we have

a4c2+a2b42a3b2c,a4b2+a2c42a3bc2,b4c2+b2a42a2b3c\dfrac {a^4c^2 + a^2b^4}{2} \geq a^3b^2c, \quad \dfrac {a^4b^2 + a^2c^4}{2} \geq a^3bc^2, \quad \dfrac {b^4c^2 + b^2a^4}{2} \geq a^2b^3c

b4a2+b2c42ab3c2,c4b2+c2a42a2bc3,c4a2+c2b42ab2c3\dfrac {b^4a^2 + b^2c^4}{2} \geq ab^3c^2, \quad \dfrac {c^4b^2 + c^2a^4}{2} \geq a^2bc^3, \quad \dfrac {c^4a^2 + c^2b^4}{2} \geq ab^2c^3

a3b3+b3c3+c3a33a2b2c22a3b3+2b3c3+2c3a36a2b2c2a^3b^3 + b^3c^3 + c^3a^3 \geq 3a^2b^2c^2 \quad \Rightarrow \quad 2a^3b^3 + 2b^3c^3 + 2c^3a^3 \geq 6a^2b^2c^2

Summing these, we get the expansion above. Thus, the statement is true.

Sharky Kesa - 3 years, 10 months ago

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Consider this:

We just need to prove (ba+b+bb+c)(cb+c+cc+a)(ac+a+aa+b)1\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\left(\dfrac{c}{b+c}+\dfrac{c}{c+a}\right)\left(\dfrac{a}{c+a}+\dfrac{a}{a+b}\right)\le 1 However by AM-GM LHS(ba+b+bb+c+cb+c+cc+a+ac+a+aa+b3)3=1LHS\le \left(\begin{array}{c}\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{c}{c+a}+\dfrac{a}{c+a}+\dfrac{a}{a+b}\\ \hline 3\end{array}\right)^3=1 A beautiful solution, not by me :P

Daniel Liu - 3 years, 10 months ago

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Problem 3
Prove the following inequality if a,b,ca,b,c are positive reals and a+b+c+abc=4a+b+c+abc=4 ,
(1+ab+ac)(1+bc+ba)(1+ca+cb)27(1+\dfrac{a}{b} +ac)(1+\dfrac{b}{c} +ba)(1+\dfrac{c}{a} +cb) \geq 27

Pakistan Round 1-2016

Sualeh Asif - 3 years, 10 months ago

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Problem 6 :

Let x>y>zx > y > z be positive reals greater than one.

Prove that cycx73>cycx2y13\large{ \sum_{cyc} x^{\frac{7}{3}} > \sum_{cyc} x^{2} y^{\frac{1}{3}}}

Harsh Shrivastava - 3 years, 10 months ago

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Solution 6: Thank You @Harsh Shrivastava,

Since we are given x,y,zR+x,y,z \in \mathbb{R^{+} } and x>y>zx>y>z.

We have two sets x2>y2>z2x^2 > y^2 > z^2 and x13>y13>z13x^{\frac{1}{3}} > y^{\frac{1}{3}} > z^{\frac{1}{3}}.

By rearrangement inequality we have

x2x13+y2y13+z3z13>x2y13+y2z13+z2x13cyclic x73> cyclic x2y13\Large{{ x }^{ 2 }{ x }^{ \frac { 1 }{ 3 } }+{ y }^{ 2 }{ y }^{ \frac { 1 }{ 3 } }+{ z }^{ 3 }{ z }^{ \frac { 1 }{ 3 } }>{ x }^{ 2 }{ y }^{ \frac { 1 }{ 3 } }+{ y }^{ 2 }{ z }^{ \frac { 1 }{ 3 } }+{ z }^{ 2 }{ x }^{ \frac { 1 }{ 3 } }\\ \therefore \quad \sum _{ \text{cyclic } }^{ }{ { x }^{ \frac { 7 }{ 3 } } } >\sum _{ \text{ cyclic } }^{ }{ { x }^{ 2 } } { y }^{ \frac { 1 }{ 3 } }}

Now my time to post questions!! Harsh By the way how did you thought of this?

Department 8 - 3 years, 10 months ago

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Problem 13

Given that a,b,ca,b,c are non-negative reals satisfying a+b+c=1a+b+c=1, prove that (1a)(1b)(1c)727+abc(1-a)(1-b)(1-c)\le \dfrac{7}{27}+abc

Daniel Liu - 3 years, 10 months ago

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Solution to Problem 13

I expanded (once more) to get a solution.

Replace (1a),(1b),(1c)(1-a), (1-b), (1-c) with (b+c),(c+a),(a+b)(b+c), (c+a), (a+b) in the in equation to get:

(a+b)(b+c)(c+a)727+abc(a+b)(b+c)(c+a) \leq \dfrac {7}{27} + abc

Expanding and simplifying, we get

a2b+a2c+b2a+b2c+c2a+c2b+abc727a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc \leq \dfrac {7}{27}

Removing the denominator, we get

27a2b+27a2c+27b2a+27b2c+27c2a+27c2b+27abc727a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7

Homogenising the RHS, we get

27a2b+27a2c+27b2a+27b2c+27c2a+27c2b+27abc7(a+b+c)327a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7(a+b+c)^3

27a2b+27a2c+27b2a+27b2c+27c2a+27c2b+27abc7a3+7b3+7c3+21a2b+21a2c+21b2a+21b2c+21c2a+21c2b+42abc27a^2b + 27a^2c + 27b^2a + 27b^2c + 27c^2a + 27c^2b + 27abc \leq 7a^3 + 7b^3 + 7c^3 + 21a^2b + 21a^2c + 21b^2a + 21b^2c + 21c^2a + 21c^2b + 42abc

6a2b+6a2c+6b2a+6b2c+6c2a+6c2b7a3+7b3+7c3+15abc6a^2b + 6a^2c + 6b^2a + 6b^2c + 6c^2a + 6c^2b \leq 7a^3+ 7b^3 + 7c^3 + 15abc

a2(b+c)+b2(a+c)+c2(a+b)76a3+76b3+76c3+52abca^2(b + c) + b^2(a + c) + c^2(a + b) \leq \dfrac {7}{6} a^3 + \dfrac {7}{6} b^3 + \dfrac {7}{6} c^3 + \dfrac {5}{2} abc

Note that by Schur's, a2(b+c)+b2(a+c)+c2(a+b)a3+b3+c3+3abca^2(b + c) + b^2(a + c) + c^2(a + b) \leq a^3 + b^3 + c^3 + 3abc. We will prove the sharper inequality:

a3+b3+c3+3abc76a3+76b3+76c3+52abca^3 + b^3 + c^3 + 3abc \leq \dfrac {7}{6} a^3 + \dfrac {7}{6} b^3 + \dfrac {7}{6} c^3 + \dfrac {5}{2} abc

6a3+6b3+6c3+18abc7a3+7b3+7c3+15abc6a^3 + 6b^3 + 6c^3 + 18abc \leq 7a^3 + 7b^3 + 7c^3 + 15abc

3abca3+b3+c33abc \leq a^3 + b^3 + c^3

Which is obviously true by AM-GM. Thus, proven.

Sharky Kesa - 3 years, 10 months ago

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Note that this problem is actually equivalent to IMO 1984 #1.

Extension: can you find the best constants for the RHS?

Daniel Liu - 3 years, 10 months ago

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Problem 5

Let xx, yy, zz be positive reals such that xyz=1xyz=1. Show that

2(x+1)2+y2+1+2(y+1)2+z2+1+2(z+1)2+x2+11\dfrac {2}{(x+1)^2+y^2+1} + \dfrac {2}{(y+1)^2+z^2+1} + \dfrac {2}{(z+1)^2+x^2+1} \leq 1

Sharky Kesa - 3 years, 10 months ago

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Solution 5 :

(xy)20(x-y)^{2} \geq 0

    x2+y22xy\implies x^{2} + y^{2} \geq 2xy

Adding 2x+22x+2 on both sides, x2+y2+2x+22xy+2x+2x^{2} + y^{2} +2x + 2 \geq 2xy +2x +2

    1xy+x+12x2+y2+2x+2\implies \dfrac{1}{xy+x+1} \geq \dfrac{2}{x^{2} + y^{2} +2x + 2 }

    1xy+x+12(x+1)2+y2+1\implies \dfrac{1}{xy+x+1} \geq \dfrac{2}{(x+1)^2+y^2+1 }

Similarly , 1yz+y+12(y+1)2+z2+1\dfrac{1}{yz+y+1} \geq \dfrac{2}{(y+1)^2+z^2+1 } and 1zx+z+12(z+1)2+x2+1\dfrac{1}{zx+z+1} \geq \dfrac{2}{(z+1)^2+x^2+1 }

Adding the three inequalities, we get 2(x+1)2+y2+1+2(y+1)2+z2+1+2(z+1)2+x2+11yz+y+1+1xy+x+1+1xz+z+1\dfrac {2}{(x+1)^2+y^2+1} + \dfrac {2}{(y+1)^2+z^2+1} + \dfrac {2}{(z+1)^2+x^2+1} \leq \dfrac{1}{yz+y+1} + \dfrac{1}{xy+x+1}+ \dfrac{1}{xz+z+1}

Since 1yz+y+1+1xy+x+1+1xz+z+1=1\dfrac{1}{yz+y+1} + \dfrac{1}{xy+x+1}+ \dfrac{1}{xz+z+1} = 1 , the result is evident.

Harsh Shrivastava - 3 years, 10 months ago

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Problem 14

Show that for non-negative real xx, yy and zz,

2x3(x3+8y3)+2y3(y3+8z3)+2z3(z3+8x3)9x4(y2+z2)+9y4(z2+x2)+9z4(x2+y2)2x^3 (x^3 + 8y^3) + 2y^3 (y^3 + 8z^3) + 2z^3 (z^3 + 8x^3) \geq 9x^4 (y^2 + z^2) + 9y^4 (z^2 + x^2) + 9z^4 (x^2 + y^2)

Sharky Kesa - 3 years, 10 months ago

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SOLUTION OF PROBLEM 14 LHSRHS=cyc(xy)4(x2+4xy+y2)0LHS - RHS = \displaystyle \sum_{cyc} (x-y)^4(x^2 + 4xy + y^2) \geq 0 Hence proved

Chris Galanis - 3 years, 10 months ago

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@Chris Galanis Can you please tell how you came up with that factorization?

A Former Brilliant Member - 3 years, 10 months ago

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@A Former Brilliant Member Actually when I solve inequalities this way I expect LHSRHS0LHS-RHS \geq 0. Since x,y,zx, y, z are non negative I expect all the negative terms to form a square. In particular I tried to find pattern like a22ab+b2a^2 - 2ab +b^2 (which is (ab)2(a-b)^2) or a44a3b+6a2b24ab3+b4a^4 - 4a^3b + 6a^2b^2 -4ab^3 + b^4 (which is (ab)4(a-b)^4). Just that

Chris Galanis - 3 years, 10 months ago

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@Chris Galanis Nice!

A Former Brilliant Member - 3 years, 10 months ago

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Problem 25

For the positive real numbers a,b,ca, b, c prove that: 1b(a+b)+1c(b+c)+1a(c+a)272(a+b+c)2\dfrac{1}{b(a+b)} + \dfrac{1}{c(b+c)} + \dfrac{1}{a(c+a)} \geq \dfrac{27}{2(a+b+c)^2}

Chris Galanis - 3 years, 10 months ago

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Solution to Problem 25

This solution is quite long and extremely bashy. Sorry. :P

We have to prove

cyc1b(a+b)272(a+b+c)2\displaystyle \sum_{\text{cyc}} \dfrac {1}{b(a+b)} \geq \dfrac {27}{2(a+b+c)^2}

Multiplying out the denominators and simplifying (here is the bash, but I have suppressed it), we get

cyc2a5b+6a4b2+2a2b4+6a3b3+8a4bccyc7a3b2c+9a3bc2+8a2b2c2\displaystyle \sum_{\text{cyc}} 2a^5 b + 6a^4 b^2 + 2a^2 b^4 + 6a^3 b^3 + 8a^4bc \geq \displaystyle \sum_{\text{cyc}} 7a^3 b^2 c + 9a^3 b c^2 + 8 a^2 b^2 c^2

By AM-GM, we have

a5b+a4b2+c4a2+a2b4+a2b4+a3b3+a3c37a21b14c77\dfrac {a^5 b + a^4 b^2 + c^4 a^2 + a^2 b^4 + a^2 b^4 + a^3 b^3 + a^3 c^3}{7} \geq \sqrt[7]{a^{21}b^{14}c^7}

a5b+a4b2+c4a2+c4a2+c4a2+a3b3+a3b3+a3c3+a3c39a27b9c189\dfrac {a^5 b + a^4 b^2 + c^4 a^2 + c^4 a^2 + c^4 a^2 + a^3 b^3 + a^3 b^3 + a^3 c^3 + a^3 c^3}{9} \geq \sqrt[9]{a^{27}b^{9}c^{18}}

8a4bc+8ab4c+8abc4324a6b6c63\dfrac {8a^4bc + 8ab^4c + 8abc^4}{3} \geq \sqrt[3]{24a^6 b^6 c^6}

These inequations imply

a5b+a4b2+c4a2+2a2b4+a3b3+a3c37a3b2ca^5 b + a^4 b^2 + c^4 a^2 + 2a^2 b^4 + a^3 b^3 + a^3 c^3 \geq 7a^3 b^2 c

a5b+a4b2+3c4a2+2a3b3+2a3c39a3bc2a^5 b + a^4 b^2 + 3c^4 a^2 + 2a^3 b^3 + 2a^3 c^3 \geq 9a^3 b c^2

8a4bc+8ab4c+8abc424a2b2c28a^4bc + 8ab^4c + 8abc^4 \geq 24a^2 b^2 c^2

Adding the first two equations cyclicly, then adding the third equation, we get the statement we wished to prove. Thus proven!

PS: It took me hours to get the right grouping of terms for the AM-GM.

Sharky Kesa - 3 years, 10 months ago

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You could have grouped them by triples of powers that majorize each other, and then applied AM-GM... Saves you all the time of finding your own grouping ;)

Daniel Liu - 3 years, 10 months ago

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PROBLEM 2 :

Let a,ba,b and cc be non-negative reals such that a+b+c=1a+b+c = 1.

Show that 1+6abc14+3(a+b)(b+c)(c+a)1 + 6abc \geq \dfrac{1}{4} + 3(a+b)(b+c)(c+a)

Harsh Shrivastava - 3 years, 10 months ago

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Homogenizing and expanding gives a3+b3+c3+6abca2b+ab2+b2c+bc2+c2a+ca2a^3+b^3+c^3+6abc \ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 which is just Schur.

Daniel Liu - 3 years, 10 months ago

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Solution Problem 2:
1+6abc14+3(1a)(1b)(1c)1+6abc \geq \dfrac{1}{4} +3(1-a)(1-b)(1-c)
1+6abc14+3(1(a+b+c)+cycababc)1+6abc \geq \dfrac{1}{4} +3(1-(a+b+c)+\sum_{cyc} ab -abc)
1+6abc14+3(cycababc)1+6abc \geq \dfrac{1}{4} +3(\sum_{cyc} ab -abc)
34+9abc3×cycab\dfrac{3}{4}+9abc \geq 3 \times \sum_{cyc} ab
14+3abccycab0\dfrac{1}{4}+3abc - \sum_{cyc} ab \geq 0
1+12abc4cycab1+ 12abc \geq 4 \sum_{cyc} ab
(a+b+c)3+12abccycab(a+b+c)^3+12abc \geq \sum_{cyc} ab
a3+b3+c3+6abcsyma2ba^3+b^3+c^3+6abc \geq \sum_{sym} a^2b
This is schurs inequality.

Sualeh Asif - 3 years, 10 months ago

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How did you use ab+bc+ca13ab+bc+ca \le \dfrac{1}{3} on 1+12abc4(ab+bc+ca)1+12abc\ge 4(ab+bc+ca) to get abc0abc\ge 0?

Daniel Liu - 3 years, 10 months ago

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Some of you might have seen this but still answer me.

Problem 7

Let a,b,ca, b, c be positive real numbers such that 1a+1b+1c=a+b+c \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=a+b+c. Prove that

1(2a+b+c)2+1(2b+c+a)2+1(2c+a+b)2316\Large{\frac { 1 }{ { \left( 2a+b+c \right) }^{ 2 } } +\frac { 1 }{ { \left( 2b+c+a \right) }^{ 2 } } +\frac { 1 }{ { \left( 2c+a+b \right) }^{ 2 } } \le \frac { 3 }{ 16 } }

Department 8 - 3 years, 10 months ago

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Solution to Problem 7

By AM-GM 1(2a+b+c)214(a+b)(a+c)\dfrac{1}{(2a+b+c)^2} \le \dfrac{1}{4(a+b)(a+c)} and thus it remains to prove cyc1(a+b)(a+c)34\sum_{cyc} \dfrac{1}{(a+b)(a+c)}\le \dfrac{3}{4} Clearing denominators, it remains to prove 8(a+b+c)3(a+b)(b+c)(c+a)8(a+b+c)\le 3(a+b)(b+c)(c+a) and after homogenizing this becomes 8abc(a+b+c)23(ab+bc+ca)(a+b)(b+c)(c+a)8abc(a+b+c)^2 \le 3(ab+bc+ca)(a+b)(b+c)(c+a) which after expansion and simplification is 3syma3b2syma3bc+2syma2b2c3\sum_{sym}a^3b^2 \ge \sum_{sym}a^3bc+2\sum_{sym}a^2b^2c which is true by Muirhead.

Daniel Liu - 3 years, 10 months ago

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Problem 10

Given that a2,a3,,ana_2, a_3, \ldots , a_n is a permutation of {2,3,,n}\{2, 3, \ldots , n\}, prove that (2a21)(3a31)(nan1)(n1)!(n+1)!2(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge \dfrac{(n-1)!(n+1)!}{2}

Daniel Liu - 3 years, 10 months ago

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Solution 10 :

Observe that(n1)!(n+1)!2=(n21)((n1)21)(321)(221)\dfrac{(n-1)!(n+1)!}{2} = (n^{2}-1)((n-1)^{2} - 1) \cdots (3^{2}-1)(2^{2}-1)

So now the inequality becomes (2a21)(3a31)(nan1)(n21)((n1)21)(321)(221)(2a_2-1)(3a_3-1)\cdots (na_n-1)\ge (n^{2}-1)((n-1)^{2} - 1) \cdots (3^{2}-1)(2^{2}-1)

This can be proven by applying Reverse Rearrangement Inequality on the sets X:(a21),(a31),(an1){X} : (a_{2}-1),(a_{3}-1) \cdots, (a_{n}-1) and Y:(1×a2),(2×a3),(3×a4),((n1)×an){Y} : (1 \times a_{2}), (2\times a_{3}),(3\times a_{4}) \cdots,( (n-1) \times a_{n}).

X,YX,Y will be similarly ordered when ak=ka_{k} = k.

So applying Reverse Rearrangement Inequality (Random Sum Product > Similar ordered Product), i=2n(Xi+Yi)i=2n(i21)\displaystyle\prod_{i=2}^n (X_{i}+Y_{i}) \geq \displaystyle\prod_{i=2}^n(i^{2} - 1)

Harsh Shrivastava - 3 years, 10 months ago

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PROBLEM 12

If a,b,c,da,b,c,d are positive reals such that abcd=1abcd=1, prove that 1(1+a)(1+a2)+1(1+b)(1+b2)+1(1+c)(1+c2)+1(1+d)(1+d2)1\dfrac{1}{(1+a)(1+a^{2})}+\dfrac{1}{(1+b)(1+b^{2})}+\dfrac{1}{(1+c)(1+c^{2})}+\dfrac{1}{(1+d)(1+d^{2})} \ge 1

A Former Brilliant Member - 3 years, 10 months ago

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Problem 15

Let x1,x2,x3,,xm>1x_1, x_2, x_3, \ldots, x_m >1 ,where mN+m \in \mathbb{N^+}, be positive integers such that xj<xj+1x_j < x_{j+1} (1j<m)(1 \leq j < m).

Prove that i=1m1xi3<1\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} < 1

Chris Galanis - 3 years, 10 months ago

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Another interesting solution.

We got xii+11xi31(i+1)3<1(i+1)2<1i(i+1)=1i1i+11xi3<1i1i+1()x_i \geq i+1 \Leftrightarrow \frac{1}{x_i^3} \leq \frac{1}{(i+1)^3} < \frac{1}{(i+1)^2} < \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1} \Leftrightarrow \frac{1}{x_i^3} < \frac{1}{i} - \frac{1}{i+1}(*) Thus taking the sum for i=1i = 1 until i=mi = m: i=1m1xi3<()i=1m1i1i+1=11m+1<1\displaystyle \sum_{i=1}^m \frac{1}{x_i^3} \stackrel{(*)}{<} \sum_{i=1}^m \frac{1}{i} - \frac{1}{i+1} = 1 - \frac{1}{m+1} <1 Since i=1m1i1i+1\displaystyle \sum_{i=1}^m \frac{1}{i} - \frac{1}{i+1} is a telescoping series. Hence proved

Chris Galanis - 3 years, 10 months ago

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i=1m1xi3<i=11x3 dx=12<1\sum_{i=1}^{m} \dfrac{1}{x_i^3} < \int_{i=1}^{\infty}\dfrac{1}{x^3}\text{ d}x = \dfrac{1}{2} < 1

Daniel Liu - 3 years, 10 months ago

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@Chris Galanis, @Sualeh Asif, @Daniel Liu, @Sharky Kesa, @Harsh Shrivastava, @Svatejas Shivakumar. Thanks everyone fit making this such a beautiful contest, hope it went on! For season 2 I will be the organiser, comment please for it.

Department 8 - 3 years, 9 months ago

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Sure!

Harsh Shrivastava - 3 years, 9 months ago

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More than welcome :)

A Former Brilliant Member - 3 years, 9 months ago

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Waiting forit to begin...

Sualeh Asif - 3 years, 9 months ago

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Problem 11 :

If a,b,ca,b,c are positive real numbers that satisfy a+b+c=1a+b+c=1, prove that a+bc+b+ac+c+ba2\sqrt{a+bc}+ \sqrt{b+ac} + \sqrt{c+ba} \leq 2

Harsh Shrivastava - 3 years, 10 months ago

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SOLUTION OF PROBLEM 11

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)=1(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=1 or ab+bc+ca13ab+bc+ca \le \dfrac{1}{3} Sincea2+b2+c2ab+bc+ca\color{#D61F06}{\text{Since}} \quad \color{#D61F06}{{a^{2}+b^{2}+c^{2} \ge ab+bc+ca}}

Now, by Cauchy-Schwarz inequality ((a+bc)2+(b+ac)2+(c+ba)2)(1+1+1)a+bc+b+ac+c+ba\sqrt{\left((\sqrt{a+bc})^{2}+(\sqrt{b+ac})^{2}+(\sqrt{c+ba})^{2} \right)(1+1+1)} \ge \sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba} or 3+3(ab+bc+ca)a+bc+b+ac+c+ba\sqrt{3+3(ab+bc+ca)} \ge \sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba}

Now 3+3(ab+bc+ca)2\sqrt{3+3(ab+bc+ca)} \le 2 since ab+bc+ca13ab+bc+ca \le \dfrac{1}{3}. Hence a+bc+b+ac+c+ba2\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ba} \le 2.

Equality holds when a=b=c=13a=b=c=\dfrac{1}{3}. Hence proved.

A Former Brilliant Member - 3 years, 10 months ago

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Can you please elaborate more on how you applied Cauchy-Schwarz here?

Chris Galanis - 3 years, 10 months ago

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@Chris Galanis It's a just applying Cauchy-Schwarz with a+bc,b+ac,c+baa+bc,b+ac,c+ba and (1,1,1) then, taking square root on both sides.

A Former Brilliant Member - 3 years, 10 months ago

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Problem 16

For positive reals a,b,ca,b,c, prove aa2+8bc+bb2+8ca+cc2+8ab1. \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1.

Source: ISL

Daniel Liu - 3 years, 10 months ago

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Solution to Problem 16

I used Jensen's Inequality to solve this problem.

Apply Jensen's on the convex function

f(x)=1xf(x) = \dfrac {1}{\sqrt{x}}

and weights aa, bb, cc in the following way.

a f(a2+8bc)+b f(b2+8ca)+c f(c2+8ab)a+b+cf(a(a2+8bc)+b(b2+8ca)+c(c2+8ab)a+b+c)\dfrac {a \text{ } f(a^2 + 8bc) + b \text{ } f(b^2 + 8ca) + c \text{ } f(c^2 + 8ab)}{a+b+c} \geq f \left (\dfrac {a(a^2 + 8bc) + b(b^2 + 8ca) + c(c^2 + 8ab)}{a+b+c} \right )

This may be rewritten as

aa2+8bc+bb2+8ca+cc2+8ab(a+b+c)32a3+b3+c3+24abc.\dfrac{a}{\sqrt{a^2 + 8bc}} + \dfrac{b}{\sqrt{b^2 + 8ca}} + \dfrac{c}{\sqrt{c^2 + 8ab}} \geq \dfrac {(a+b+c)^{\frac {3}{2}}}{\sqrt {a^3 + b^3 + c^3 + 24abc}}.

Thus, it suffices to prove

(a+b+c)3a3+b3+c3+24abc(a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc

But we can expand this and simplify to get

a2b+a2c+b2a+b2c+c2a+c2b+2abc8abca^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc \geq 8abc