Maths Olympiad Prep

Track / Stage 6 / 194 of 400 #1194 of 1964

Problem 1194

National olympiad, first round
Algebra Difficulty 6.3 Prove it

4.2.41 Let a1,a2,⋯ ,ana_{1}, a_{2}, \cdots, a_{n} be nn positive numbers (n⩾2)(n \geqslant 2), not all equal, and satisfying ∑k=1nak−2n\sum_{k=1}^{n} a_{k}^{-2 n} =1=1. Prove that: ∑k=1nak2n−n2∑1⩽i<j⩽nai−2aj−2>n2\sum_{k=1}^{n} a_{k}^{2 n}-n^{2} \sum_{1 \leqslant i<j \leqslant n} a_{i}^{-2} a_{j}^{-2} > n^{2}.

The text to be translated into English, please retain the original text's line breaks and format, and output the translation result directly.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

For any x>0x>0, by the arithmetic-geometric mean inequality, we have ∑i=0n−1xn−1−2i⩾n\sum_{i=0}^{n-1} x^{n-1-2 i} \geqslant n, with equality holding if and only if x=1x=1. Also, (x−x−1)(∑i=0n−1xn−1−2i)=xn−x−n\left(x-x^{-1}\right)\left(\sum_{i=0}^{n-1} x^{n-1-2 i}\right)=x^{n}-x^{-n}. Therefore,
(xn−x−n)2=(x−x−n)2(∑i=0n−1xn−1−2i)2⩾n2(x−x−1)2, so ∑k=1nak∑k=1n1ak−n2=∑1⩽i<j⩽n(aiaj2n−ajai2n)=∑1⩽i<j⩽n[(aiaj2n)n−(ajai2n)n]2=∑1⩽i<j⩽n(xijn−xij−n)( where xij=aiaj2n)⩾n2∑1⩽i<j⩽n(xij−xij−1)2=n2∑1⩽i<j⩽n(aiaj2najai2n)2, \begin{array}{l} \left(x^{n}-x^{-n}\right)^{2}=\left(x-x^{-n}\right)^{2}\left(\sum_{i=0}^{n-1} x^{n-1-2 i}\right)^{2} \geqslant n^{2}\left(x-x^{-1}\right)^{2}, \text { so } \sum_{k=1}^{n} a_{k} \sum_{k=1}^{n} \frac{1}{a_{k}}-n^{2} \\ \quad=\sum_{1 \leqslant i<j \leqslant n}\left(\sqrt[2 n]{\frac{a_{i}}{a_{j}}}-\sqrt[2 n]{\frac{a_{j}}{a_{i}}}\right)=\sum_{1 \leqslant i<j \leqslant n}\left[\left(\sqrt[2 n]{\frac{a_{i}}{a_{j}}}\right)^{n}-\left(\sqrt[2 n]{\frac{a_{j}}{a_{i}}}\right)^{n}\right]^{2} \\ \quad=\sum_{1 \leqslant i<j \leqslant n}\left(x_{i j}^{n}-x_{i j}^{-n}\right)\left(\text { where } x_{i j}=\sqrt[2 n]{\frac{a_{i}}{a_{j}}}\right) \\ \geqslant n^{2} \sum_{1 \leqslant i<j \leqslant n}\left(x_{i j}-x_{i j}^{-1}\right)^{2}=n^{2} \sum_{1 \leqslant i<j \leqslant n}\left(\sqrt[2 n]{\frac{a_{i}}{a_{j}}} \sqrt[2 n]{\frac{a_{j}}{a_{i}}}\right)^{2}, \end{array}

with equality holding if and only if a1=a2=⋯=ana_{1}=a_{2}=\cdots=a_{n}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.