# 1995 IMO Problems/Problem 2

## 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/

### Solution 8 (fast Titu's Lemma no substitutions)

Rewrite as .

Now applying Titu's lemma yields .

Now applying the AM-GM inequality on . The result now follows.

Note: , because . (Why? Because , and hence ).

~th1nq3r

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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.*