TITLE: Proving $\pi^3 \gt 31$
QUESTION [6 upvotes]: $$\large \pi^3 \gt 31$$
Using a calculator, $\pi^3/31 \approx 1.0002$, so I thought this may be challenging to do by hand.
It is extremely easy with the use of any calculator, so I was wondering now:

Can you prove the above inequality without the use of calculator or advanced computation in an elegant manner?

REPLY [2 votes]: From 
$$\sum_{k=0}^\infty \frac{960}{(2k+1)^6} = \pi^6$$
we have 
$$\sum_{k=1}^\infty \frac{960}{(2k+1)^6} = \pi^6-960$$
and
$$\sum_{k=2}^\infty \frac{960}{(2k+1)^6} = \pi^6-961-\frac{77}{243}$$
Since
$$960<961<961+\frac{77}{243}$$
we can form a new series for $\pi^6-961$ as a weighted sum of the two truncations. 
Solving the equation
$$\left(\pi^6-960\right)a+\left(\pi^6-961-\frac{77}{243}\right)b=\pi^6-961$$
for rational $a$ and $b$ yields
$$a=\frac{77}{320}$$
$$b=\frac{243}{320}$$
Finally, 
$$\pi^6-961=(\pi^3-31)(\pi^3+31)=3\sum_{k=0}^\infty \left(\frac{77}{(2k+3)^6}+\frac{243}{(2k+5)^6}\right)$$
so 

$$\pi^3-31=\frac{3}{\pi^3+31}\sum_{k=0}^\infty \left(\frac{77}{(2k+3)^6}+\frac{243}{(2k+5)^6}\right)$$

is positive because the series contains only positive terms.