# 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 ?

• You can start by showing that $\Gamma (w+ 1) > 1 - \gamma w \Leftrightarrow \Gamma (w) > \frac{1}{w} - \gamma$ for $0<w<1$. Then $$\Gamma \left( {\sin \left( {\frac{1}{x}} \right)} \right) \ge \frac{1}{{\sin \left( {\frac{1}{x}} \right)}} - \gamma \ge \frac{1}{{\frac{1}{x}}} - \gamma = x - \gamma$$ for $x\geq 1$. Something similar should work for the other term.
– Gary
Commented Apr 20, 2021 at 11:02
• @Gary First thanks for the hint but how come up the $\frac{1}{3}$ for the second term ? Commented Apr 20, 2021 at 16:12
• It seems to me that $$- \Gamma \left( {\sin ^2 \left( {\tfrac{1}{x}} \right)} \right) + x^2 < \gamma - \tfrac{1}{3}$$ for $x\geq 1$, so, unfortunately, my proposed bound would not be enough to conclude.
– Gary
Commented Apr 20, 2021 at 17:16

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.