# Weird Looking Inequality

Here is a very weird looking inequality I think I have found. Any ideas about how to prove it are much appreciated!

For any constants $a>0$ and $c>0$ we have:

$\frac{x^2-1}{2}+\ln \left(x\right) \leq -\frac{1}{2}\ln \left(1-\left(\frac{x^2-1}{cx}\right)^a\right)+\frac{c^2}{2}\left(\frac{x^{2\ }-1}{cx}\right)^{\left(2-a\right)}$

which holds for all $x \geq 1$. For large enough parameters $a,c$ there seems to be a single point of equality with $x>1$

Here is a link to the plots of these functions so you can play around with the parameters $a$ and $c$. https://www.desmos.com/calculator/vxujknshe6

• Very beautiful question (+1) – Ahmad Sep 20 '17 at 23:34
• A very slight modification of your formula: there is an unnecssary factor $1/2$. – Jean Marie Sep 21 '17 at 6:29

It turns out the inequality can be manipulated to proving that for $c>0,k>0,x>0$ that
$$\sqrt{c}x^{k/2}\leq\sinh\left(x+cx^{k-1}-\sqrt{c}x^{k/2}\right)$$
The original result then follows by doing some substitution and using the formula for $\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1})$. This easier inequality is proven by AM-GM: start with $\sqrt{2x\cdot2cx^{k-1}}\leq \frac{1}{2}\left(2x+2cx^{k-1}\right)$, then subtract $\sqrt{c}x^{k/2}$ from both sides, and finally use $x < \sinh(x)$.