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
5 years, 3 months ago

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1 vote

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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 - 5 years, 1 month ago

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Waiting forit to begin...

Sualeh Asif - 5 years, 1 month ago

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More than welcome :)

A Former Brilliant Member - 5 years, 1 month ago

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Sure!

Harsh Shrivastava - 5 years, 1 month ago

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@Sharky Kesa The contest ended? Wish it could have continued for some for time. Anyway thanks to everyone who participated in this contest, learnt a lot and had a lot of fun.

A Former Brilliant Member - 5 years, 1 month ago

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Problem 28

For the real numbers a,b,c,da, b, c, d prove that: 1ac+bda2+b2c2+d21-1\le \dfrac{ac+bd}{\sqrt{a^2+b^2}\sqrt{c^2+d^2}} \le 1

Chris Galanis - 5 years, 2 months ago

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For the right hand side note that for positive real numbers(or all negative) (a2+b2)(c2+d2)(ac+bd)2(a^2 +b^2)(c^2+d^2)\geq (ac+bd)^2 By C.S .

Now it suffices to consider the case when one of a,b,c,da,b,c,d are negative. If one of them is negative. WLOG assume aa then let A=aA=-a. Thus (bdAc)2(bd+ac)2(a2+b2)(c2+d2)(bd-Ac)^2\leq (bd+ac)^2 \leq (a^2+b^2)(c^2+d^2) equality holds when A=a=0A=a=0

If any two of them are negative we have two cases .

WLOG assume a,ba,b. Then let A=aA=-a, B=bB=-b, (Ac+Bd)2=(ac+bd)2\therefore (Ac+Bd)^2=(ac+bd)^2 Otherwise assume both of a,ca,c are negative.Then there multiple is positive and we are done.

The case when 3 are negative is equivalent to the case when one is negative.

Sualeh Asif - 5 years, 2 months ago

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Yet that is not a proof since a,b,c,da, b, c, d may not be positive

Chris Galanis - 5 years, 2 months ago

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@Chris Galanis Of course completed the solution.

I tend to keep posting my progress because the mobile app gets stuck

Sualeh Asif - 5 years, 2 months ago

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Problem 27

Given positive reals a,b,ca,b,c, prove that 3cyca+bc10+8a2+b2+c2ab+bc+ca3\sum_{cyc}\dfrac{a+b}{c}\ge 10+8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}

Daniel Liu - 5 years, 2 months ago

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(1) (a+b)(b+c)(c+a)89(a+b+c)(ab+bc+ca)(a+b)(b+c)(c+a) \ge \dfrac{8}{9}(a+b+c)(ab+bc+ca) which is true since by expanding and simplifying we get syma2b6abc\displaystyle \sum_{sym} a^2b \ge 6abc which is rearrangement inequality.

(2) (syma2b)+2abc=(a+b)(b+c)(c+a)\Big(\displaystyle \sum_{sym} a^2b\Big)+2abc = (a+b)(b+c)(c+a) and (a+b+c)2=a2+b2+c2+2(ab+bc+ca)(a+b+c)^2 = a^2+b^2+c^2 +2(ab+bc+ca)


3cyca+bc10+8a2+b2+c2ab+bc+ca3 \displaystyle \sum_{cyc} \dfrac{a+b}{c} \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca} Adding 66 to both sides and using (2) we get:

3(a+b)(b+c)(c+a)abc8(a+b+c)2ab+bc+ca3\dfrac{(a+b)(b+c)(c+a)}{abc} \ge 8\dfrac{(a+b+c)^2}{ab+bc+ca} Thus by (1) it suffices to prove that:

(ab+bc+ca)23abc(a+b+c)(ab+bc+ca)^2\ge 3abc(a+b+c) Which is true since by expanding and simplifying both sides we get rearrangement inequality. Hence proved!

Chris Galanis - 5 years, 2 months ago

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Problem 26

Given positive reals xx, yy, zz greater than or equal to 1 such that

1x+1y+1z=2\dfrac {1}{x} + \dfrac {1}{y} + \dfrac {1}{z} = 2

Prove

x+y+zx1+y1+z1\sqrt{x+y+z} \geq \sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1}

Sharky Kesa - 5 years, 2 months ago

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Another solution:

The problem is equivalent to (x+y+z)(31x1y1z)(x1+y1+z1)2(x+y+z)\left(3-\dfrac{1}{x}-\dfrac{1}{y}-\dfrac{1}{z}\right)\ge \left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2

Confirm that 11x0    x11-\dfrac{1}{x}\ge 0\implies x\ge 1; thus, by C-S we are done.

Daniel Liu - 5 years, 2 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 - 5 years, 2 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 - 5 years, 2 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 - 5 years, 2 months ago

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Sorry for the delay guys, and the inconvenience caused by me.

Problem 24:

Let a,b,c be positive reals.

Prove that cyc(1+ab)2(1+a+b+c(abc)13)\displaystyle \prod_{cyc} (1 + \dfrac{a}{b}) \geq 2(1 + \dfrac{a+b+c}{ (abc)^{\frac{1}{3}}})

Harsh Shrivastava - 5 years, 3 months ago

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Open up the brackets and simplify a bit to get cyc(ab+ac)2(a+b+c)(abc)13\large \sum_{cyc} (\frac{a}{b} + \frac{a}{c}) \geq 2\dfrac{(a+b+c)}{(abc)^{\frac{1}{3}}}

Substitute a=x3,b=y3andc=z3a = x^{3}, b= y^{3} and c = z^{3}, we get cyc(x3y3+x3z3)2(x3+y3+z3)xyz=cyc(x2yz+x2yz)\large \sum_{cyc} (\frac{x^{3}}{y^{3}} + \frac{x^{3}}{z^{3}}) \geq 2\dfrac{(x^{3} + y^{3}+ z^{3})}{xyz} = \sum_{cyc} (\dfrac{x^{2}}{yz} + \dfrac{x^{2}}{yz})

Now using re-arrangement inequality, we can get our desired result.

Harsh Shrivastava - 5 years, 3 months ago

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Hmm sorry guyz I won't be able to post solution now because I am a bit busy. Will post a solution by tmmrw evening(my time zone).

Harsh Shrivastava - 5 years, 3 months ago

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I think this is equivalent to (1+ab)(1+bc)(1+ca)2(1+a+b+c(abc)1/3)(1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a}) \geq 2(1+\dfrac{a+b+c}{(abc)^{1/3}})

By applying Holder's Inequality to LHS it suffices to prove that ((1+(abbcca)1/3)32(1+a+b+c(abc)1/3)41+a+b+c(abc)1/3(abc)1/3a+b+c3\Big((1+(\dfrac{a}{b} \dfrac{b}{c} \dfrac{c}{a})^{1/3}\Big)^3 \geq 2(1+\dfrac{a+b+c}{(abc)^{1/3}}) \\ \Rightarrow 4 \geq 1+\dfrac{a+b+c}{(abc)^{1/3}} \\ \Rightarrow (abc)^{1/3} \geq \dfrac{a+b+c}{3} But by AM-GM this is true only for a=b=c=1a=b=c=1


Edit: If the problem had the direction of the inequality flipped then we would have to prove that

(1+ab)(1+bc)(1+ca)2(1+a+b+c(abc)1/3)AMGM2(1+a+b+ca+b+c/3)=8 \begin{aligned} (1+\dfrac{a}{b})(1+\dfrac{b}{c})(1+\dfrac{c}{a}) & \leq 2(1+\dfrac{a+b+c}{(abc)^{1/3}}) \\ & \stackrel{AM-GM}{\leq}2(1+\dfrac{a+b+c}{{a+b+c}/_3}) = 8\end{aligned}

Then it suffices to prove that (a+b)(b+c)(c+a)8abcAMGM(23(a+b+c))38abca3+b3+c3+3a2b+3a2c+3ab2+3ac2+3b2c+3bc221abc \begin{aligned} & (a+b)(b+c)(c+a) \leq 8abc \\ & \stackrel{AM-GM}{\leq} \Big(\dfrac{2}{3}(a+b+c)\Big)^3 \leq 8abc \\ & \Rightarrow a^3 + b^3 +c^3 +3a^2b+3a^2c+3ab^2+3ac^2+3b^2c+3bc^2 \leq 21abc \end{aligned} But again that would be true by rearrangement inequality if the direction of the inequality was flipped...

Chris Galanis - 5 years, 3 months ago

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The problem is that every time you apply an inequality, the inequality weakens a bit. If you apply inequalities in the incorrect fashion or too much, then the original inequality will weaken to the point that it is false.

Daniel Liu - 5 years, 2 months ago

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Next problem please!

A Former Brilliant Member - 5 years, 2 months ago

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@A Former Brilliant Member I request someone to post a problem on my behalf. Thanks!

Harsh Shrivastava - 5 years, 2 months ago

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@Chris Galanis I hope you post your problem soon!

Sualeh Asif - 5 years, 3 months ago

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@Sualeh Asif This was not a solution... I wait for Harsh or someone else to post the solution

Chris Galanis - 5 years, 3 months ago

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@Sualeh Asif I believe that it isn't a solution. The sign must be interchanged. Even I got that after applying A.M-G.M. You pretty much cannot use anything I guess. Such a strange inequality.

A Former Brilliant Member - 5 years, 3 months ago

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@A Former Brilliant Member Since the inequality is homogenous, we assume abc=1abc=1.
We can factor it to cyc(a+b)(ab1)0\sum_{cyc} (a+b)(ab-1)\geq 0

Sualeh Asif - 5 years, 3 months ago

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Problem 23

Find the greatest positive real number kk, for which is true that xy(x2+y2)(3x2+y2)1k\dfrac{xy}{\sqrt{(x^2+y^2)(3x^2+y^2)}} \leq \dfrac{1}{k} for all positive real numbers x,yx, y.

Chris Galanis - 5 years, 3 months ago

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Harsh Shrivastava - 5 years, 3 months ago

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Problem 22

Here is a very nice problem I came across.

Given that a,b,ca,b,c are non-negative reals, prove that (a2+2)(b2+2)(c2+2)3(a+b+c)2(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2

Daniel Liu - 5 years, 3 months ago

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Solution of Problem 22

By Cauchy we got: (a2+2)(1+(b+c)22)(a+b+c)2(a^2+2)(1+\frac{(b+c)^2}{2}) \geq (a+b+c)^2

Then it suffices to prove that (b2+2)(c2+2)3(1+(b+c)22)(b^2+2)(c^2+2) \geq 3(1+\frac{(b+c)^2}{2})

By rearranging and simplifying we get: (b22bc+c2)+(2b2c24bc+1)0(bc)2+2(bc1)20(b^2-2bc+c^2) + (2b^2c^2-4bc+1) \geq 0 \\ (b-c)^2 + 2(bc-1)^2 \geq 0 Which is true. Equality holds iff a=b=c=1a=b=c=1.

Chris Galanis - 5 years, 3 months ago

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Can you please elaborate on your first step? I don't seem to see the Cauchy. Did you perhaps mean (b+c)22\dfrac{(b+c)^2}{2} instead of b+c2\dfrac{b+c}{2}?

EDIT: I tried to solve the problem using your idea and it appears that you have only just typoed. Splendid solution!

Daniel Liu - 5 years, 3 months ago

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@Daniel Liu Sorry, I fixed that !

Chris Galanis - 5 years, 3 months ago

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Problem 21

If for the real numbers a,b,ca, b, c where bc0bc \neq 0 is true that 1c2bc0\frac{1-c^2}{bc} \geq 0, prove that: 10(a2+b2+c2bc3)2ab+5ac10(a^2+b^2+c^2-bc^3) \geq 2ab + 5ac

Chris Galanis - 5 years, 3 months ago

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If c1|c| \ge 1 and sgn(b)=sgn(c)\text{sgn}(b) = -\text{sgn}(c), then the inequality is 10(a2+b2+c2bc3)10(a2+b2+c2)2ab+5ac10(a^2+b^2+c^2-bc^3)\ge 10(a^2+b^2+c^2)\ge 2ab+5ac    (ab)2+52(ac)2+132a2+9b2+152c20\iff (a-b)^2+\dfrac{5}{2}(a-c)^2+\dfrac{13}{2}a^2+9b^2+\dfrac{15}{2}c^2\ge 0 solved.

If c1|c| \le 1 and sgn(b)=sgn(c)\text{sgn}(b) = \text{sgn}(c), then the inequality is 10(a2+b2+c2bc3)10(a2+b2+c2bc)2ab+5ac10(a^2+b^2+c^2-bc^3)\ge 10(a^2+b^2+c^2-bc)\ge 2ab+5ac    (ab)2+52(ac)2+5(bc)2+132a2+4b2+52c20\iff (a-b)^2+\dfrac{5}{2}(a-c)^2+5(b-c)^2+\dfrac{13}{2}a^2+4b^2+\dfrac{5}{2}c^2\ge 0 done.

Not a very strong inequality; no equality case in fact.

Daniel Liu - 5 years, 3 months ago

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Can this problem be done without calculus? I have no idea how to proceed with this problem.

A Former Brilliant Member - 5 years, 3 months ago

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Me and Sharky were discussing about it and it showed this can be nearly solved using x20,xRx^2 \ge 0, x\in \mathbb{R}.

Department 8 - 5 years, 3 months ago

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Problem 20

Let aa, bb and cc be positive reals such that abc=8abc=8. Prove that

a2(1+a3)(1+b3)+b2(1+b3)(1+c3)+c2(1+c3)(1+a3)43.\dfrac {a^2}{\sqrt{(1+a^3)(1+b^3)}} + \dfrac {b^2}{\sqrt{(1+b^3)(1+c^3)}} + \dfrac {c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \dfrac {4}{3} .

Sharky Kesa - 5 years, 3 months ago

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Solution of Problem 20

Note that 1+k3=(1+k)(1k+k2)AMGMk2k+1+k+12=k2+222k2+21(1+k3)()\sqrt{1+k^3} = \sqrt{(1+k)(1-k+k^2)} \stackrel{AM-GM}{\leq} \frac{k^2-k+1+k+1}{2} = \frac{k^2+2}{2} \Leftrightarrow \frac{2}{k^2+2} \leq \frac{1}{\sqrt{(1+k^3)}} (*) Hence: cyca2(1+a3)(1+b3)43()cyc4a2(a2+2)(b2+2)433cyca2(c2+2)(a2+2)(b2+2)(c2+2)cyc2a2+cyca2b272 \begin{aligned} \displaystyle \sum_{cyc} \frac{a^2}{\sqrt{(1+a^3)(1+b^3)}} & \geq \frac{4}{3} \\ \stackrel{(*)}{\Rightarrow} \sum_{cyc} \frac{4a^2}{(a^2+2)(b^2+2)} & \geq \frac{4}{3} \\ \Rightarrow 3 \sum_{cyc} a^2(c^2+2) & \geq (a^2+2)(b^2+2)(c^2+2) \\ \Rightarrow \sum_{cyc} 2a^2+ \sum_{cyc} a^2b^2 & \geq 72 \end{aligned} Which is true by applying AM-GM in both sums separately. Hence proved.

Chris Galanis - 5 years, 3 months ago

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Problem 19 :

Let x,y,zx,y,z be positive real numbers.Prove that cyc1x3+y3+xyz1xyz \displaystyle \sum_{cyc} \dfrac{1}{x^{3} + y^{3} + xyz} \leq \dfrac{1}{xyz}

Harsh Shrivastava - 5 years, 3 months ago

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Solution to Problem 19

Since this inequality is homogenous, WLOG xyz=1xyz = 1. We will prove that

1x3+y3+1zx+y+z\dfrac {1}{x^3+y^3+1} \leq \dfrac {z}{x+y+z}

(x3+y3)z+zx+y+z(x^3+y^3)z+z \geq x+y+z

Note that x3+y3=(x+y)(x2xy+y2)x^3+y^3 = (x+y)(x^2-xy+y^2).

(x+y)(x2xy+y2)zx+y(x+y)(x^2-xy+y^2)z \geq x+y

(x2+y2xy)z1(x^2 + y^2 - xy)z \geq 1

By AM-GM, x2+y22xyx^2 + y^2 \geq 2xy. Substituting, we get

(2xyxy)z1(2xy-xy)z \geq 1

xyz1xyz \geq 1

Which is obviously true by our assumption. Thus, proven. The inequality in the question can be achieved by adding cyclicly.


We have already proven (with WLOG xyz=1xyz=1 because of the equation's homogeneity)

1x3+y3+1zx+y+z\dfrac {1}{x^3 + y^3 + 1} \leq \dfrac {z}{x+y+z}

This means that

1x3+y3+1+1y3+z3+1+1z3+x3+1x+y+zx+y+z\dfrac {1}{x^3 + y^3 + 1} + \dfrac {1}{y^3 + z^3 + 1} + \dfrac {1}{z^3 + x^3 + 1} \leq \dfrac {x+y+z}{x+y+z}

cyc1x3+y3+11\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + 1} \leq 1

Note that since xyz=1xyz=1, 1xyz=1\dfrac {1}{xyz} = 1. Thus,

cyc1x3+y3+xyz1xyz\displaystyle \sum_{\text{cyc}} \dfrac {1}{x^3 + y^3 + xyz} \leq \dfrac {1}{xyz}

Sharky Kesa - 5 years, 3 months ago

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PROBLEM 18

For positive real numbers a,b,ca,b,c with a+b+c=abca+b+c=abc, prove that 11+a2+11+b2+11+c232\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.

A Former Brilliant Member - 5 years, 3 months ago

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Solution Of Problem 18 :

Substitute a= tanx , b= tany , c= tanz such that x+y+z=π x+y+z= \pi

Since a,b,c are positive reals, x,y,z are all less than π2 \frac{\pi}{2}

Now we want to prove that cosx+cosy+cosz1.5 \cos x + \cos y + \cos z \leq 1.5

By Jensen's Inequality,

cos(x+y+z3)cosx+cosy+cosz3\cos \left(\dfrac{x+y+z}{3}\right)\ge \dfrac{\cos x+\cos y+\cos z}{3}

    cosx+cosy+cosz32 \implies \cos x + \cos y + \cos z \leq \dfrac{3}{2}

Note that f(p)=cospf(p) = \cos p is concave down down for all 0pπ2 0 \leq p \leq \frac{\pi}{2} .This can be checked by differentiating f(p) twice.

Harsh Shrivastava - 5 years, 3 months ago

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Problem 17

For positive reals xx, yy and zz such that xyz=1xyz=1, prove that

1x3(y+z)+1y3(z+x)+1z3(x+y)32\dfrac {1}{x^3(y+z)} + \dfrac {1}{y^3(z+x)} + \dfrac {1}{z^3 (x+y)} \geq \dfrac {3}{2}

Sharky Kesa - 5 years, 3 months ago

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This is a very famous inequality that appeared in IMO in the late 90s I believe. My approach goes as follows.

SOLUTION OF PROBLEM 17

Dividing the numerator and denoninator by x2,y2x^2,y^2 and z2z^2 respectively and applying Titu's Lemma we get 1x3(y+z)+1y3(z+x)+1z3(x+y)(1x+1y+1z)22(xy+yz+zx)=(xy+yz+zx)22(xy+yz+zx)=xy+yz+zx2\dfrac {1}{x^3(y+z)} + \dfrac{1}{y^3(z+x)}+\dfrac{1}{z^3 (x+y)} \geq \dfrac{(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2}{2(xy + yz+ zx)}=\dfrac{(xy+yz+zx)^2}{2(xy+yz+zx)}=\dfrac{xy+yz+zx}{2}

Now, by A.M-G.M inequality xy+yz+zx23(xyz)232=32\dfrac{xy+yz+zx}{2} \geq 3 \cdot \dfrac{\sqrt[3]{(xyz)^2}}{2}=\dfrac{3}{2}.

Equality holds if and only if x=y=z=1x=y=z=1. Hence proved.

A Former Brilliant Member - 5 years, 3 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 - 5 years, 3 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

(a+b)(b+c)(c+a)8abc(a+b)(b+c)(c+a) \geq 8abc

Which is true by AM-GM. Thus proven.

Sharky Kesa - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 months ago

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

A Former Brilliant Member - 5 years, 3 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 - 5 years, 3 months ago

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@Chris Galanis Nice!

A Former Brilliant Member - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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 - 5 years, 3 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