"The tetrahedron described on the sphere of radius r has the smallest (smallest) area (volume) when it is regular." - Is that true?
Because I want to solve that problem:
In a tetrahedron, prove that: $$(\frac{1}{6}.S_{tp})^3 \geq \sqrt{3}.V^2$$ $S_{tp}$ is total area of four triangles of that tetrahedron. V is the volume of that tetrahedron.
Not sure what you mean "described on ... for smallest area/volume when it is regular".
However,
Part I - proof of the inequality.
Given any $n$ vectors $u_1, u_2, \ldots, u_n$ in $\mathbb{R}^N$, its Gram matrix $[u_i\cdot u_j]_{i,j=1}^n$ is a $n\times n$ matrix with entries $u_i \cdot u_j$. Its determinant can be expressed in terms of the norm of the exterior product of vectors $u_i$.
$$G(u_1,\ldots,u_n) \stackrel{def}{=} \det[u_i\cdot u_j]_{i,j=1}^n = \| u_1 \wedge \cdots\wedge u_n \|^2$$
In particular, for $n = 1, 2, 3$, in $\mathbb{R}^3$, this reduces to
$$G(u_1) = |u_1|^2,\quad G(u_1,u_2) = |u_1 \times u_2|^2\quad\text{ and }\quad G(u_1,u_2,u_3) = |u_1 \cdot ( u_2 \times u_3 )|^2$$
The characteristic polynomial of a Gram matrix has following decomposition:
$$\chi_{[u_i\cdot u_j]}(\lambda) \stackrel{def}{=}\det(\lambda I_n - [u_i\cdot u_j]_{i,j=1}^n) = \sum_{k=0}^N (-1)^k\lambda^{N-k} \sum_{1\le i_1 < \cdots < i_k \le n} G(u_{i_1},\cdots,u_{i_k})$$
For four vectors $u_1, \ldots, u_4$ in $\mathbb{R}^3$, this becomes:
$$\chi_{[u_i\cdot u_j]}(\lambda) = \lambda( \lambda_3 - c_1\lambda^2 + c_2 \lambda + c_3) \quad\text{ where }\quad \begin{cases} c_1 &= \sum_{i=1}^4 |u_i|^2\\ c_2 &= \sum_{1\le i < j\le 4} |u_i \times u_j|^2\\ c_3 &= \sum_{1\le i < j < k \le 4} |u_i \cdot ( u_j \times u_k )|^2 \end{cases}$$ If $u_1, u_2, u_3, u_4$ are not coplanar, at least one of $|u_i \cdot (u_j \times u_k)| \ne 0$ and $c_3 > 0$. Since a Gram matrix is always positive semi-definite, the $4\times 4$ matrix $[u_i\cdot u_j]_{i,j=1}^4$ will have one zero and three positive roots. Apply Maclaurin's inequality to these three positive roots, we obtain $$\frac{c_1}{3} \ge \left(\frac{c_2}{3}\right)^{1/2} \ge c_3^{1/3}\tag{*1}$$
Given any non-degenerate tetrahedron with vertices $v_1, v_2, v_3, v_4$ with volume $V$. Let $\vec{S}_k = S_k\hat{n}_k$ where $S_k$ is the area and $\hat{n}_k$ is the outward unit normal vector of the face opposite to $v_k$.
One can show that
$$ |\vec{S}_1 \cdot (\vec{S}_2 \times \vec{S}_3)| = |\vec{S}_2 \cdot (\vec{S}_3 \times \vec{S}_4)| = |\vec{S}_3 \cdot (\vec{S}_4 \times \vec{S}_1)| = |\vec{S}_4 \cdot (\vec{S}_1 \times \vec{S}_2)| = \frac{9}{2}V^2$$
If one substitute $u_k$ by $\hat{n}_k = \frac{\vec{S}_k}{S_k}$ in $(*1)$, $\frac{c_1}{3} \ge c_3^{1/3}$ becomes
$$\begin{align} &\frac43 \ge \left\{\left(\frac{9}{2}V^2\right)^2\left[ \frac{1}{(S_1S_2S_3)^2} + \frac{1}{(S_2S_3S_4)^2} + \frac{1}{(S_3S_4S_1)^2} + \frac{1}{(S_4S_1S_2)^2} \right]\right\}^{1/3}\\ \implies & (S_1S_2S_3S_4)^2 \ge \frac{3^7}{4^4}V^4(S_1^2+S_2^2+S_3^2+S_4^2) \ge \frac{3^7}{4^3}V^4 (S_1S_2S_3S_4)^{1/2}\\ \iff & (S_1S_2S_3S_4)^{3/4} \ge \left(\frac{3}{2}\right)^3\sqrt{3} V^2 \end{align} $$ From this, we can deduce:
$$\left(\frac16 S_{tp}\right)^3 = \left(\frac16 (S_1+S_2+S_3+S_4)\right)^3 \ge \left(\frac23(S_1S_2S_3S_4)^{1/4}\right)^3 \ge \sqrt{3}V^2\tag{*2} $$
Above proof is constructed based on information I extracted from a chinese book 四面体不等式 (Inequality in the the Tetrahedron) by 樊益武. If you have access to the book and if you can read chinese, look at $\S2.3$, 四面体顶点角 for more background materials.
Part II - regular (and only regular) tetrahedrons have smallest volume.
WOLOG, we only need to look at those tetrahedron with inradius $r = 1$.
Since $V = \frac13 S_{tp}r$, $(*2)$ implies for any tetrahedron with unit inradius,
$$\left(\frac{V}{2}\right)^3 \ge \sqrt{3}V^2 \quad\implies\quad V \ge 8\sqrt{3}$$ Since regular tetrahedrons with inradius $1$ has volume $8\sqrt{3}$, their volume is indeed the smallest.
On the other direction, if a tetrahedron with unit inradius has volume $8\sqrt{3}$. $(*2)$ tell us the AM and GM of $S_1, S_2, S_3, S_4$ coincides. This forces $$S_1 = S_2 = S_3 = S_4 = \frac{S_{tp}}{4}$$ Notice for any tetrahedron, we have the vector identity: $$\vec{S}_1 + \vec{S}_2 + \vec{S}_3 + \vec{S}_4 = S_1 \hat{n}_1 + S_2 \hat{n}_2 + S_3\hat{n}_3 + S_4\hat{n}_4 = \vec{0}$$ This means for any tetrahedron with smallest volume, $$\hat{n}_1 + \hat{n}_2 + \hat{n}_3 + \hat{n}_4 = 0 \quad\implies\quad \sum_{i=1}^4\sum_{j=1}^4 \hat{n}_i\cdot\hat{n}_j = 0 \quad\implies\quad \sum_{i < j} \hat{n}_i\cdot\hat{n}_j = -2 $$
Using this, we can express the coefficient of $c_2$ in $(*1)$ as,
$$\begin{align}c_2 &= \sum_{i < j}|\hat{n}_i \times \hat{n}_j|^2 = \sum_{i < j}\left[1 - (\hat{n}_i \cdot \hat{n}_j)^2\right] = \sum_{i < j}\left[\frac{10}{9} - (\hat{n}_i \cdot \hat{n}_j + \frac13)^2 + \frac23 \hat{n}_i\cdot\hat{n}_j \right]\\ &= \frac{16}{3} - \sum_{i < j}(\hat{n}_i \cdot \hat{n}_j + \frac13)^2 \end{align} $$ There is another requirement for $(*2)$ to be true, the inequalities in $(*1)$ need to become equality. This places another constraint on $c_2$:
$$\frac{c_1}{3} = (c_3)^{1/3} \implies c_2 = 3\left(\frac{c_1}{3}\right)^2 = \frac{16}{3}$$ Compare the two expression for $c_2$, we find:
$$\sum_{i < j}(\hat{n}_i \cdot \hat{n}_j + \frac13)^2 = 0 \quad\iff\quad \hat{n}_i \cdot \hat{n}_j = -\frac13 \;\;\text{ for all } i \ne j $$ Since the faces of the tetrahedron are planes of the form $\hat{n}_k\cdot \vec{x} = 1$. Above condition determines the relative "orientations" among all four faces uniquely. This force any tetrahedron with smallest volume $8\sqrt{3}$ to be a regular one.