, 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:
$$\begin{aligned} \frac{y^{2n} - 1}{y - 1} - \frac{n y^{n-1}(y^2 - 1)}{y - 1} &= \sum_{i=0}^{2n-1} y^i - n(y^{n-1} + y^n) = \\ &= \sum_{i=0}^{n-1} (y^{2n-1-i} - y^n - y^{n-1} + y^i) = \sum_{i=0}^{n-1} y^i (y^{n-1-i} - 1)(y^{n-i} - 1) \ge 0. \end{aligned}$$