It's a follow up of Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. :
Let $x>4/\pi$ then we have :
$$\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\cos\left(\frac{2}{x}\right)>0$$
My strategy :
For $x\geq 2$ we have :
$$\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)\geq 2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\geq \left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)^{\frac{1}{\sqrt{x}}}$$
Remains to show :
$$\lim_{n\to \infty}\cos\left(\frac{2}{x^{2n}}\right)-\left(\frac{x^{2n}-1}{x^{2n}+1}\right)^{\frac{1}{x^{2n}}}=0$$
Lemma 1 :
Let $x\geq 1$ $$\frac{\left(x^{2}-1\right)}{x^{2}+1}-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$
Proof :
For the RHS we have (using the same approach as in my answer):
$$f(x)=1+\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)+\frac{1}{2}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)^{2}+\frac{1}{6}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{1}{x}-2\right)\left(\frac{x-1}{x+1}-1\right)^{3}\geq\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$
And :
$$f(x)-\frac{x^{2}-1}{x^{2}+1}=-\frac{2\left(x-1\right)\left(x^{3}-2x^{2}+7x-2\right)}{3\left(x^{2}+x\right)^{3}\left(x^{2}+1\right)}\leq 0$$
Now we need to show for $x\geq 2$:
$$\frac{x^{4}-1}{x^{4}+1}-2\frac{x^{2}-1}{x^{4}+\frac{2}{3}+\frac{1}{3}x^{2}}-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}\geq 0$$
But for $x\geq 2$: $$\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}\geq g(x)=\left(\frac{x-1}{x+1}\right)^{-1}\left(1+\left(\frac{1}{x}+1\right)\left(\frac{x-1}{x+1}-1\right)+\frac{1}{2}\left(\frac{1}{x}+1\right)\frac{1}{x}\left(\frac{x-1}{x+1}-1\right)^{2}+\frac{1}{6}\left(\frac{1}{x}-1\right)\frac{1}{x}\left(\frac{1}{x}+1\right)\left(\frac{x-1}{x+1}-1\right)^{3}\right)$$
But :
$$\frac{x^{4}-1}{x^{4}+1}-2\frac{x^{2}-1}{x^{4}+\frac{2}{3}+\frac{1}{3}x^{2}}-g(x)=\frac{-2\left(3x^{8}-3x^{7}-x^{6}-3x^{5}+4x^{4}-6x^{3}-4x^{2}-6x+4\right)}{(3x^{3}(x+1)(x^{4}+1)(3x^{4}+x^{2}+2))}<0$$
So the RHS is shown for $x\ge 2 $
Now the LHS :
$$h(x)=\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)-2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|$$
We have for $x\in[0,\pi/2]$:
$$1-\frac{x^{2}}{2}+\frac{x^{4}}{24}\geq \cos(x)\geq 1-\frac{x^{2}}{2}$$
So for $x\geq 4 $ we have :
$$r\left(\frac{2}{x}\right)-t\left(\frac{2}{\sqrt{x}}\right)-2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\leq 0$$
Where :
$$t\left(x\right)=1-\frac{x^{2}}{2}+\frac{x^{4}}{24},r\left(x\right)=1-\frac{x^{2}}{2}$$
So we are done for $x\geq 4$.
Have you another (simpler) proof ?
Bonus/conjecture :
Let $x>1$ such that $x\in[\alpha,\beta]$ then define :
$$m\left(x\right)=\cos\left(\frac{2}{x}\right)-\left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}$$
Then it seems $\exists a,b,c\in[-\infty,\infty]$ and $0<\varepsilon$ arbitrary small such that :
$$b^{-x^{a}}+c-\varepsilon<\sum_{n=1}^{\infty}\left(-1\right)^{n}m\left(x^{n}\right)<b^{-x^{a}}+c$$
All the goal is to determine $\alpha,\beta$