I would like to prove $\zeta(11)>\pi^3/31$, where $\zeta$ is the Riemann zeta function. They differ in the fourth decimal place. I suppose this could be 'proven' by simply comparing known numerical approximations for the two constants, which are accurate beyond the fourth decimal place, but that is no fun.
The first step would be to show whether $\pi^3 <31$ or $\pi^3 > 31$ while hoping $\pi^3 \neq 31$. It is not too difficult to prove the second, so I am only asking for help proving $\zeta(11) > \pi^3/31$ with the assumption that $\pi^3/31>1$.
I initially hoped this inequality would spring from some other properties of the zeta function, though odd powers of $\pi$ are hard to come by. One can show that $$-\frac{2^{11}\pi^{10}\zeta'(-10)}{3628800} = \zeta(11)$$ Which is lovely and probably useless to solving my problem. I tried to use the fact that $\zeta(x)$ is strictly decreasing on the domain I am concerned with, though I have not made much progress as I cannot say much about $\zeta(11)$ by itself as there is no known closed form, and manipulating $\zeta(12)$ and $\zeta(10)$ to obtain the $\pi^3$ term leaves me $\pi$ multiples of the zeta function, which I have no idea how to handle.
Actually, since $\zeta(2k)$ is a rational multiple of $\pi^{2k}$, it would perhaps be more natural to consider that $\zeta(2k+1)>c \pi^{2k}$ for a rational $c>0$, and not $\pi^3$.
So, for $k=5$ this would be, say, $$ \zeta(11)> \frac{\pi^{10}}{93648}. $$
And as you said, we could use the following formula here.
$$ \zeta(2k+1)=\frac{(-1)^k2^{2k+1}}{(2k)!}\pi^{2k}\zeta'(-2k). $$