How prove $(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}\leq 2(1+n(x-1))^{n}$ for $n\in\mathbb{N}$? Let $x\ge 1$. How prove that $(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}\leq 2(1+n(x-1))^{n}$ for $n\in\mathbb{N}$?

The easy way:

Let $f(x)=RHS-LHS$.
 
$f'(x)=2n^2(1+n(x-1))^{n-1}-n(1+\frac{x}{\sqrt{x^2-1}})(x+\sqrt{x^2-1})^{n-1}-n(1-\frac{x}{\sqrt{x^2-1}})(x-\sqrt{x^2-1})^{n-1}$

$f'(x)=2n^2(1+n(x-1))^{n-1}-\frac{n}{\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}+\frac{n}{\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$

$f'(x) > 2n^2(1+n(x-1))^{n-1}-\frac{n^2}{x+\sqrt{x^2-1}}(x+\sqrt{x^2-1})^{n}-\frac{n^2}{x-\sqrt{x^2-1}}(x-\sqrt{x^2-1})^{n}$.

That, by induction, gives $f'(x) \geq 0$, so we are done.

Verifying that $\frac{n}{\sqrt{x^2-1}}<\frac{n^2}{x+\sqrt{x^2-1}}$, for $n \geq 2, x>1$, and formalising the inductive argument, are exercises left to the reader.
