# Difference between revisions of "1995 IMO Problems/Problem 2"

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+ | === Solution 8 (AM-GM only)=== | ||

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+ | We are given that <math>a,b,</math> and <math>c</math> are positive real numbers such that <math>abc=1.</math> This means that by AM-GM, we have <cmath>\frac{a+b+c}{3} \geq \sqrt[3]{abc}=\sqrt[3]{1},</cmath> so <math>a+b+c>3.</math> Let <math>a+b+c=3+k</math> for some nonnegative real number <math>k.</math> By AM-GM, we also have <cmath>\frac{a+(b+c)}{2} \geq \sqrt{a(b+c)}</cmath> <cmath>\frac{3+k}{2} \geq \frac{3}{2} \geq \sqrt{a(b+c)}.</cmath> Squaring the last two parts of this inequality chain, we have <cmath>a(b+c) \leq \frac{3}{2} \leq \frac{9}{4}.</cmath> Manipulating the first two parts of this inequality chain gives <cmath>\frac{2}{3} \leq \frac{1}{a(b+c)}.</cmath> <cmath>\frac{1}{a^3(b+c)} \geq \frac{2}{3a^2}.</cmath> Analogously, we obtain the two inequalities <math>\frac{1}{b^3(a+c)} \geq \frac{2}{3b^2}</math> and <math>\frac{1}{c^3(a+b)} \geq \frac{2}{3c^2}.</math> Now by AM-GM, we have <cmath>\frac{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}{3} \geq \sqrt[3]{\frac{1}{a^2 \cdot b^2 \cdot c^2}}=1,</cmath> so <math>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \geq 3.</math> Using all of our information, we can conclude that <cmath>\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)} \geq \frac{2}{3a^2}+\frac{2}{3b^2} + \frac{2}{3c^2} = \frac{2}{3} \left (\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right) \geq \frac{2}{3} \cdot 3 \geq \frac{3}{2},</cmath> and the result follows. <math>\mathbb{Q.E.D.}</math> | ||

== Resources == | == Resources == |

## Revision as of 14:58, 19 June 2021

## Contents

## Problem

(*Nazar Agakhanov, Russia*)
Let be positive real numbers such that . Prove that

## Solution

### Solution 1

We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.

### Solution 2

We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.

### Solution 3

Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.

### Solution 3b

Without clever notation: By Cauchy-Schwarz,

Dividing by and noting that by AM-GM gives as desired.

### Solution 4

After the setting and as so concluding

By Titu Lemma, Now by AM-GM we know that and which concludes to

Therefore we get

Hence our claim is proved ~~ Aritra12

### Solution 5

Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.

### Solution 6

Without clever substitutions, and only AM-GM!

Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that

### Solution 7 from Brilliant Wiki (Muirheads) =

https://brilliant.org/wiki/muirhead-inequality/

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*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*

### Solution 8 (AM-GM only)

We are given that and are positive real numbers such that This means that by AM-GM, we have so Let for some nonnegative real number By AM-GM, we also have Squaring the last two parts of this inequality chain, we have Manipulating the first two parts of this inequality chain gives Analogously, we obtain the two inequalities and Now by AM-GM, we have so Using all of our information, we can conclude that and the result follows.