Official Solution: From the QM-HM inequality for the six numbers $ |P_i-P_j|$ we have \[ \sum_{i\neq j}\frac{1}{|P_i-P_j|}\geq C\left(\sum_{i\neq j}|P_i-P_j|^2\right)^{-\frac 12}\]where $ C$ is a constant. The equality holds iff $ |P_i-P_j|$'s are all equal, which is satisfied when $ P_i$'s are the vertices of a regular tetrahedron. We now try to find an upper bound for $ \sum_{i\neq j}|P_i-P_j|^2$. Considering $ P_i$'s as vectors in 3D space, then this sum becomes $ 3\sum_i(P_i.P_i)-2\sum_{i\neq j}(P_i.P_j)$. Adding and subtracting $ \sum_i(P_i.P_i)$ we get $ 4\sum_i(P_i.P_i)-(\sum_iP_i).(\sum_iP_i)$. But $ P_i.P_i$ is a constant. Therefor the expression is maximized when $ \sum_iP_i=0$ for which it is equal to $ 4$. This condition is again satisfied when $ P_i$'s are the vertices of a regular tetrahedron. Therefore, the condition of minimality of $ \sum_{i\neq j}\frac 1{|P_i-P_j|}$ becomes that, $ |P_i-P_j|$'s should be equal for different $ i,j$'s and that $ \sum_iP_i=0$. But if $ |P_i-P_j|$ has a constant value, then every face of the tetrahedron is an equilateral triangle, which itself shows that the tetrahedron is regular.

Omid Hatami wrote: Official Solution: ... We now try to find an upper bound for $ \sum_{i\neq j}|P_i - P_j|^2$. Considering $ P_i$'s as vectors in 3D space, then this sum becomes $ 3\sum_i(P_i.P_i) - 2\sum_{i\neq j}(P_i.P_j)$... I don't see this; isn't it $ 2\left(\sum_i|P_i|^2-\sum P_i\cdot P_j\right)$? I guess it doesn't really matter in your proof. Mine was similar, except I used Cauchy engel form: \[ \sum (P_i-P_j)^{-1} \leq \frac{C}{\sum |P_i-P_j|}\] so we need upper bound for $ \left(\sum |P_i-P_j|\right)^2\leq C' \sum |P_i-P_j|^2$ then end same as above

Suppose $P_i=(\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}})$ for all $i\in\{1,2,3,4\}$. Now call $f(i,j)=|P_i-P_j|$. Certainly we need to minimize $\sum \frac {1}{f(i,j)}$. $\sum f^2(i,j)=\sum(\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}-\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2+\sum (\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}}-\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2+\sum (\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2$. And now so $\sum f^2(i,j)=4-\sum (\sum \frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2$. Now, $(\sum f^2(i,j))(6)\geq (\sum f(i,j))^2\geq (\frac {36}{\sum \frac {1}{f(i,j)}})^2$- (**) Now so $\sum \frac {1}{f(i,j)}$ is minimized when $\sum \frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=0$. Now so clearly we've two cases for when $\sum \frac {1}{f(i,j)}$ can be minimized, one if all of $P_i$ lies on a circle and vertices of a square, and if not then vertices of a regular tetrahedron. If first case would br true then, equality wouldn't hold for the inequality of (**). Thus we must have those are vertices of regular tetrahedron since $f(i,j)$ is constant for any $i,j$.

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