Tetrahedron's sphere

Let $r_1, r_2, r_3,$ and $r_4$ denote the inradii of the faces of the tetrahedron and $R$ denote the inradius of the tetrahedron.

We have the following inequality,

$\frac{1}{r_1^2}$ $+$ $\frac{1}{r_2^2}$ $+$ $\frac{1}{r_3^2}$ $+$ $\frac{1}{r_4^2}$ $\leq$ $\frac{2}{R^2}$

with equality occurring if and only if the tetrahedron is a regular one.

Let $a$ be the length of an edge of the tetrahedron. We need to find the inradius of an equilateral triangle with side length $a$ .

$r_1 = \frac{K}{s}$ where $K$ is the area of the triangle and $s$ is the semiperimeter.

$r_1 = r_2 = r_3 = r_4 = \frac{\frac{a^2\sqrt3}{4}}{\frac{3a}{2}}$ $= \frac{a\sqrt3}{6}$

Hence, $\frac{1}{r_1^2}$ $+$ $\frac{1}{r_2^2}$ $+$ $\frac{1}{r_3^2}$ $+$ $\frac{1}{r_4^2}$ $=$ $4\times(\frac{12}{a^2}$ ) $= \frac{48}{a^2}$ = $\frac{2}{R^2}$

Then, $R^2 = \frac{a^2}{24}$ and $R = \frac{a}{2\sqrt6}$

The volume of the sphere is,

$\frac{4}{3}\pi R^3 =$ $\frac{4}{3}\pi (\frac{a}{2\sqrt6})^3 = \frac{4a^3}{144\sqrt6}\pi = \frac{a^3}{36\sqrt6}\pi = \pi \sqrt6$

Hence, $a^3 = 216$ , $\boxed{a=6}$

$r=inradiuse,~~~~~~~~s=edge/side,\\ incircle~volume=\frac 4 3 *\pi*r^3=\sqrt6*\pi,\\ \therefore~\frac 4 3*r^3=\sqrt6,~~\implies~r=\sqrt[3]{\sqrt6*\dfrac 3 4}.\\ But~~also~r=\frac 1 {12}*\sqrt6*s.\\ \therefore~~s=\sqrt[3]{\sqrt6*\dfrac 3 4 }*\dfrac{12}{\sqrt6}=\sqrt[6]{6*\dfrac 9{16}}*\dfrac{12}{\sqrt6}\\ =\sqrt[6]{\dfrac{3*9} 8}*\dfrac{12}{\sqrt6}=\sqrt{\dfrac 3 2}*\dfrac{12}{\sqrt6}= \Large~~\color{#D61F06}{6}.\\$