Proving $\pi^3 \gt 31$ $$\large \pi^3 \gt 31$$
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?

 A: It is well known, at least, I know it well :-) that
$$\pi>3+\frac1{7+\dfrac1{15}}=3+\frac{15}{106}\ .$$
If you are prepared to accept this, then the rest can certainly be done with $\color{red}{\hbox{mental arithmetic}}$ (or pencil and paper at worst).  By the binomial theorem we have
$$\pi^3>3^3+3\frac{3^2\times15}{106}+3\frac{3\times15^2}{106^2}
  =27+\frac{405}{106}+\frac{45^2}{106^2}\ .$$
Now by a familiar trick,
$$45^2=2025>1900+120>1900+6\times19=106\times19$$
and so
$$\pi^3>27+\frac{405}{106}+\frac{19}{106}=31\ .$$
Comment.  The initial inequality is in fact well known to those who know about $\color{red}{\hbox{continued fractions}}$.  If it's not familiar in that context, one can quite easily divide mentally to confirm that $\frac{15}{106}<0.14151<\pi-3$.
A: The answer is given for this question
Is there an integral for $\pi^4-\frac{2143}{22}$?
Linked to this paper.
http://www.jstor.org/stable/27642693
Extract of what you want:
For $\pi^3$
The truncated continued fractions are
$31, 4930/159, 14821/478, \dots$.
\begin{align}
\pi^3-31
&=
\int_0^1
\frac{8 \, x^5 \, (1-x)^2 \, \left(324889-120736 \, x^2\right)}
{445625 \, (1 + x^2) }
\log^2 x \, dx\\
\frac{4930}{159}-\pi^3
&=
\int_0^1
\frac{4 \, x^{10} \, (1-x)^4 \, \left(695774836+470936528857 \, x^2\right)}
{470240754021 \, (1 + x^2) }
\log^2 x \, dx.
\end{align}
A: 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.)
A: 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. 
A: 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.
A: 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?
