(2014北约自主招生)已知正实数$x_1,x_2,\cdots,x_n$满足$x_1x_2\cdots x_n=1,$求证:
$(\sqrt{2}+x_1)(\sqrt{2}+x_2)\cdots(\sqrt{2}+x_n)\ge(\sqrt{2}+1)^n$
分析:根据$\dfrac{\sum\limits_{k=1}^n\dfrac{\sqrt{2}}{\sqrt{2}+x_k}}{n}\ge\sqrt[n]{\prod\limits_{k=1}^n\dfrac{\sqrt{2}}{\sqrt{2}+x_k}}=\dfrac{\sqrt{2}}{\sqrt[n]{\prod\limits_{k=1}^n(\sqrt{2}+x_k)}}$
$\dfrac{\sum\limits_{k=1}^n\dfrac{x_k}{\sqrt{2}+x_k}}{n}\ge\sqrt[n]{\prod\limits_{k=1}^n\dfrac{x_k}{\sqrt{2}+x_k}}=\dfrac{1}{\sqrt[n]{\prod\limits_{k=1}^n(\sqrt{2}+x_k)}}$
两式相加即得.
原文：https://www.cnblogs.com/mathstudy/p/10485351.html