For the tetrahedron $ABCD$ denote $AB=a$ , $BC=b$ , $CD=c$ , $DA=$ , $AC=e$ and $BD=f$ . Also let $r_A$ , $r_B$ , $r_c$ and $r_D$ be the inradii of the triangles (faces of the tetrahedron) $BCD$ , $CDA$ , $DAB$ and $ABC$ respectively . For each of these triangles apply the known inequality of Mitrinovic i.e. $\boxed{\ 3r\sqrt 3\ \le\ s\ }$ , where $s$ and $r$ are the semiperimeter and the inradius of a given triangle . Indeed, $\begin{array}{ccc} 3\sqrt 3\cdot\left(r_A+r_B+r_C+r_D\right)\ \le\ \frac 12\cdot\left[\left(b+c+f\right)+\left(c+d+e\right)+\left(d+a+f\right)+\left(a+b+e\right)\right]=a+b+c+d+e+f\end{array}$ and since $a+c=b+d=e+f=1$ , the last inequality finally reduces to : $r_A+r_B+r_C+r_D\ \le\ \frac {_3}{^{3\sqrt 3}}=\frac {_{\sqrt 3}}{^3}$ . Equality holds iff the triangles $BCD$ , $CDA$ , $DAB$ and $ABC$ are equilateral i.e when $ABCD$ is a regular tetrahedron .

We have \[[ABc]=rs=\sqrt{s(s-a)(s-b)(s-c)}.\] By AM-GM, we have \[\sqrt[3]{(s-a)(s-b)(s-c)}\le\frac{s-a+s-b+s-c}{3}=\frac{s}{3}.\] Therefore \[rs=s\sqrt{(s-a)(s-b)(s-c)}{s}=\sqrt{3}\left(\sqrt[3]{(s-a)(s-b)(s-c)}\right)^{\frac{3}{2}}\le\sqrt{s}\sqrt{\frac{s^3}{27}},\] which implies that for any triangle we have \[r\le\frac{s\sqrt{3}}{3}.\] Now applying this to each face of the tetrahedron $r_A, r_B, r_C, r_D$ yields \[\sum r_A\le\frac{1}{3\sqrt{3}}\sum s_A\] where $s_A, s_B, s_C, s_D$ denote the semiperimeters of the faces. Since each side is by exactly $2$ triangles, we have $\sum s_A=\sum\text{ sides of the tetrahedron}=3$. Hence \[r_A+r_B+r_C+r_D\le\frac{\sqrt{3}}{3}\] which equality when $s_A=s_B=s_C=s+D$. This follows from AM-GM, where $s-a=s-b=s-c$ and the triangle is equilateral, so the tetrahedron is regular.