Hi it's an inequality found by chance using an integral analogue of the Kantorovitch inequality .
The problem :
$$300+\left(\int_{0}^{10}2-x^{-x}dx\right)\cdot\left(1+e^{-1}\right)^{-2}\cdot4\left(e^{-1}\right)>100\pi$$
I already know that the integral of $f(x)=x^{-x}$ between $0$ towards $\infty$ is less than $2$ and you can found other inequalities here Charming approximation of $\pi$: $2\left(\frac{1}{2}\right)^{\phi/2}+2< \pi$, where $\phi$ is the golden ratio and here Showing $\sqrt{\frac{e}{2}}\cdot\frac{e}{\pi}\left(\frac{e}{2}-\frac{1}{e}\right)<1$ without a calculator
Edit we can use the approximation :
$$2-x^{-x}\simeq 2-\left(g\left(x^x\right)\right)^{-1}$$
Where :
$$g(x)=1+\frac{\left(\ln\left(ax\right)\right)^{c}-\left(\ln\left(a\right)\right)^{c}}{\left(\ln\left(2a\right)\right)^{c}-\left(\ln\left(a\right)\right)^{c}}$$
Where we need some constraint as $g'(2)-1=0$ and $a=10000$
For further informations see Prove or disprove: $1+\frac{\ln^c(ax)-\ln^ca}{\ln^c(2a)-\ln^ca}-x\leq 0$, where $1<x$, $a>2$, and $c$ is chosen so that $f'(2)=0$
My question :
How to show it by hand without a calculator ?
Thanks.
