Inequality $\Gamma\left(\sin\left(\frac{1}{x}\right)\right)-\Gamma\left(\sin^2\left(\frac{1}{x}\right)\right)-x+x^2+\frac{1}{3}\geq 0$

Question

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 ?

Answer 1

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.