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 $n\in\mathbb{N}$?
 A: 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.
A: As pointed out by r9m in the comments, by setting $x=\cosh t$ the inequality is equivalent to:
$$ \cosh(n t)\leq \left(\cosh t+(n-1)(\cosh t-1)\right)^n \tag{1}$$
or to:
$$ \cosh(n t)\leq \left(1+2n\sinh^2\frac{t}{2}\right)^n \tag{2}$$
or to:
$$ 1+2\sinh^2(nz)\leq \left(1+2n\sinh^2 z\right)^n \tag{3}$$
or to:
$$ \frac{1}{n}\log\left(1+2\sinh^2(nz)\right)\leq \log\left(1+2n\sinh^2 z\right) \tag{4}.$$
Equality is achieved in $z=0$; by considering the difference of the derivatives it is sufficient to prove that:
$$ (n-1)\sinh(nu)\geq n\sinh((n-1)u).\tag{5}$$
Equality is achieved in $u=0$, by differentiating again we are left to prove that:
$$ \cosh(nu)\geq \cosh((n-1)u) \tag{6}$$
that is trivial.
