<s>[INST] Inequality $\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x+x^2+\frac{1}{3}\geq 0$ Problem found with the help of Desmos and my imagination .
Let $x\geq 1$ then we have :
$$\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x+x^2+\frac{1}{3}\geq 0$$
Some informations :
We have the limits :
$$\lim_{x\to \infty}\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-x=-\gamma$$
And :
$$\lim_{x\to \infty}\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x^2=-\gamma+\frac{1}{3}$$
It doesn't work with higher degree like $3$.It seems to be a little  mysterious .
The derivative is very complicated and I don't expose it here .
How to show the inequality ?
 [/INST] Some thoughts
Let $f(v) = \Gamma(v) - \frac{1}{v}$.
Fact 1: $f'(v) > 0$ for all $v$ in $(0, 1)$.
Now, by Fact 1, we have
$$\Gamma(\sin(1/x)) - \frac{1}{\sin (1/x)}
\ge \Gamma(\sin^2(1/x)) - \frac{1}{\sin^2(1/x)}.$$
Thus, it suffices to prove that
$$\frac{1}{\sin (1/x)} - \frac{1}{\sin^2(1/x)} - x + x^2 + \frac{1}{3}\ge 0$$
or
$$- \left(\frac{1}{\sin (1/x)} - \frac{1}{2}\right)^2 - x + x^2 + \frac{7}{12} \ge 0.$$
Fact 2: $1/2 < \frac{1}{\sin (1/x)} \le \frac{60 x^3 + 3x}{60x^2 - 7}$ for all $x \ge 1$.
By Fact 2, it suffices to prove that
$$- \left(\frac{60 x^3 + 3x}{60x^2 - 7} - \frac{1}{2}\right)^2 - x + x^2 + \frac{7}{12} \ge 0$$
which is true.
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