It's a curious inequality which I am struggling with :
Prove that :
$$\int_{0}^{\infty}\left(e^{-\frac{t^{6}}{3+\sqrt{3}}}\right)dt<\zeta(3)$$
For information and using WA we have :
$$(3+\sqrt{3})^{1/6}\Gamma(7/6)=\int_{0}^{\infty}\left(e^{-\frac{t^{6}}{3+\sqrt{3}}}\right)dt$$
We have also a representation according to Wiki for The Apery's constant which is :
$$\frac{1}{2}\int_{0}^{\infty}\frac{x^{2}}{e^{x}-1}dx$$
An attempt :
We can compare :
$$\frac{1}{100^{2}}e^{-\frac{(\frac{x}{100^{2}})^{6}}{3+\sqrt{3}}}\cdot\frac{(6e^{-\frac{(\frac{x}{100})^{6}}{3+\sqrt{3}}}(\frac{x}{100})^{5})}{3+\sqrt{3}}2\cdot1.2020563,\frac{1}{2a}\frac{\left(\frac{x}{a}\right)^{2}}{e^{\frac{x}{a}}-1},a=100$$
Question :
Despite the accuracy can we pretend to show it analytically ?