1995 IMO Problems/Problem 2
(Nazar Agakhanov, Russia) Let be positive real numbers such that . Prove that
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.
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.
Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.
Without clever notation: By Cauchy-Schwarz,
Dividing by and noting that by AM-GM gives as desired.
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
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.
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) =
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 ).
Scroll all the way down Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.