(2014北约自主招生)已知正实数$x_1,x_2,cdots,x_n$满足$x_1x_2cdots x_n=1,$求证:
$(sqrt{2}+x_1)(sqrt{2}+x_2)cdots(sqrt{2}+x_n)ge(sqrt{2}+1)^n$
分析:根据$dfrac{sumlimits_{k=1}^ndfrac{sqrt{2}}{sqrt{2}+x_k}}{n}gesqrt[n]{prodlimits_{k=1}^ndfrac{sqrt{2}}{sqrt{2}+x_k}}=dfrac{sqrt{2}}{sqrt[n]{prodlimits_{k=1}^n(sqrt{2}+x_k)}}$
$dfrac{sumlimits_{k=1}^ndfrac{x_k}{sqrt{2}+x_k}}{n}gesqrt[n]{prodlimits_{k=1}^ndfrac{x_k}{sqrt{2}+x_k}}=dfrac{1}{sqrt[n]{prodlimits_{k=1}^n(sqrt{2}+x_k)}}$
两式相加即得.