Cesko-Slovacko-Poljsko 2013

Without loss of generality assume that $y=\sqrt{x}>1$. The identity $$a^2+\frac{1}{a^2}-2={\left(a-\frac{1}{a}\right)}^2$$ reduces the problem to showing that $$y^n-\frac{1}{y^n}\ge n\left(y-\frac{1}{y}\right)$$ or $y^{2n}-n(y^{n+1}-y^{n-1})-1\ge 0$. Upon division by $y-1>0$ this follows from the following computation: $$\begin{array}{rl}\frac{y^{2n}-1}{y-1}-\frac{n{y}^{n-1}(y^2-1)}{y-1}& =\sum _{i=0}^{2n-1}{y}^i-n(y^{n-1}+y^n)=\\ & =\sum _{i=0}^{n-1}(y^{2n-1-i}-y^n-y^{n-1}+y^i)=\sum _{i=0}^{n-1}{y}^i(y^{n-1-i}-1)(y^{n-i}-1)\ge 0.\end{array}$$