Proving $\pi^3 \gt 31$

The simplest way is to use the series:

$$\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}+...=\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^6}=\frac{\pi^6}{960}$$

Now we want to prove that $$\frac{\pi^6}{960}>\frac{31^2}{960}$$ which means that we need to prove

$$960\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^6}>31^2=961$$

However it is

$$960\sum\limits_{k=1}^{\infty}\frac{1}{(2k-1)^6}>960\left(\frac{1}{1^6}+\frac{1}{3^6}\right)=960+\frac{960}{3^6}$$

And now $\frac{960}{3^6}>1$ meaning $960>3^6$ because $320>3^5=9^2 \cdot 3=81 \cdot 3 = 243$

(Very minimal calculations involved, just to prove the point, although it was almost obvious once we got there.)

We know that $\zeta(6)=\frac{\pi^6}{945}$ and $\sum_{k=1}^\infty \frac1{k^6}$ converges quite quickly. It turns out we only need $k=1,2,3$ to prove the inequality.

$$\zeta(6)=\frac{\pi^6}{945}\Leftrightarrow \pi^3=\sqrt{945\zeta(6)}\Rightarrow \pi^3>\sqrt{945\sum_{k=1}^3 \frac1{k^6}}=\frac{\sqrt{4982145}}{72}>31.$$

Calculating the last square root is not very easy but it proves the inequality.

Well $\pi^3>3.1415^3>31$. Cubing 3.1415 is not pleasant by hand but most certainly can be done. Or are you looking for a more elegant solution?