Problem :
Show the following inequality :
$$\int_{0}^{\infty}x^{-4\left(x^{2}-x-1\right)}dx> \int_{0}^{1}x^{-x}dx$$
My attempt :
I recall the equality :
$${\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}\\\end{aligned}}}$$
Next I use a Reversed Cauchy Schwarz inequality see Reverse Cauchy Schwarz for integrals :
$$\frac{\sqrt{\int_{1}^{\frac{1+\sqrt{5}}{2}}x^{-8\left(x^{2}-x-1\right)}dx\cdot\int_{1}^{\frac{1+\sqrt{5}}{2}}1dx}}{\frac{1}{2}\left(\sqrt{\frac{f\left(1\right)}{f\left(1.315\right)}}+\sqrt{\frac{f\left(1.315\right)}{f\left(1\right)}}\right)}\leq\int_{1}^{\frac{1+\sqrt{5}}{2}}x^{-4\left(x^{2}-x-1\right)}dx$$
Wich is almost sufficient to show it but almost is not exactly .
Another path should be (sufficient to show inequality) :
$$\int_{\frac{10}{9}}^{\frac{1+\sqrt{5}}{2}}x^{-4\left(x^{2}-x-1\right)}dx-\exp\left(\int_{1}^{2}\ln\left(f\left(x\right)\right)dx\right)>0$$
Where :
$$f\left(x\right)=x^{4\left(-x^{2}+x+1\right)}$$
Here I'm stuck .
Brute forcing :
we have the following inequalities :
Let $1\leq x\leq 6/5$ then we have :
$$\frac{x^{4}\left(1-2\left(x^{2}-x\right)\left(x-1\right)\right)}{1+2\left(x^{2}-x\right)\left(x-1\right)}\leq x^{-4\left(x^{2}-x-1\right)}$$
Let $1/5\leq x\leq 1$ then (it seems) we have $a=-\frac{1}{\sqrt{3}}$ :
$$\frac{x^{4}\left(1-2\left(x^{2}-x+\frac{a}{2}\right)\left(x-1\right)\right)}{1+2\left(x^{2}-x-\frac{a}{2}\right)\left(x-1\right)}\leq x^{-4\left(x^{2}-x-1\right)}$$
It seems we have also for $3/2\leq x\leq 9/5$ and $b=2$:
$$f\left(x+\frac{77}{1000}\right)<x^{-4\left(x^{2}-x-1\right)}$$
Where ($b=2$) :
$$f\left(x\right)=\frac{9}{10}\left(x^{1-b}\left(1+\left(x-1\right)\left(x^{2}-b\right)+\frac{1}{2}\left(x-1\right)^{2}\left(\left(x^{2}-b\right)\left(x^{2}-1-b\right)\right)\right)^{-1}\left(1+\left(x-1\right)x+\left(x-1\right)^{2}x\left(x-1\right)\right)\right)^{4}$$
We are near from the true .
How to show it ?
