<s>[INST] If $x\geq 1$, then $x^{x^{\cos(x)}}\geq (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$ Hi it's a curious problem that I really don't understand :

Let $x\geq 1$ then we have :
  $$x^{x^{\cos(x)}}\geq (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$$

It's very curious for me because the accuracy of the inequality increase when $x$ increase
But clearing the expression or taking both side logarithm we get :
$$x^{\cos(x)}\ln(x)\geq \ln(1-\sqrt{x}+x)(x+1)^{\cos(x)}$$
But it doesn't help at all to explain this mystery .
It seems that there is similarity between the graphic of $f(x)=x^n$ $n$ a natural number and $g(x)=x^{x^{\cos(x)}}- (1-\sqrt{x}+x)^{(x+1)^{\cos(x)}}$
I think furthermore that our function is quasiperiodic 
but it's too fuzzy as definition .
That's all for me If you have nice ideas it would be great .
Thanks a lot for sharing your time and knowledge .
 [/INST] Alternative solution:
It suffices to prove that
$$x^{\cos x}\ln x \ge (x+1)^{\cos x}\ln (1 - \sqrt{x} + x)$$
which is written as
$$ \ln x
\ge \left(1 + \frac{1}{x}\right)^{\cos x}\ln (1 - \sqrt{x} + x). $$
It suffices to prove that
$$ \ln x
\ge \left(1 + \frac{1}{x}\right)\ln (1 - \sqrt{x} + x). $$
It suffices to prove that
$$ x\ln x
\ge (x+1)\ln (1 - \sqrt{x} + x). $$
Let $f(x) = x\ln x - (x+1)\ln (1 - \sqrt{x} + x)$. We have
\begin{align}
f'(x) &= \ln \frac{x}{1 - \sqrt{x} + x} - \frac{x-1}{2(1 - \sqrt{x} + x)\sqrt{x}}\\
&\ge \frac{2(\frac{x}{1 - \sqrt{x} + x}-1)}{\frac{x}{1 - \sqrt{x} + x}+1}
 - \frac{x-1}{2(1 - \sqrt{x} + x)\sqrt{x}}\\
&= \frac{(2\sqrt{x} - 1)(\sqrt{x} - 1)^3}{2(1 - \sqrt{x} + x)\sqrt{x}(2x + 1 - \sqrt{x})}\\
&\ge 0
\end{align}
where we have used $\ln u \ge \frac{2(u-1)}{u+1}$ for all $u \ge 1$ (Note:
$\frac{\mathrm{d}}{\mathrm{d} u}(\ln u - \frac{2(u-1)}{u+1}) = \frac{(u-1)^2}{u(u+1)^2}$).
Also, $f(1) = 0$. Thus, $f(x)\ge 0$ for all $x\ge 1$.
We are done.
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