# inequality involving the product of lengths of edges of a tetrahedron

Is there a constant $C$ such that for every tetrahedron with edges $a,b,c,d,e,f$ and volume $V$, the following inequality holds: $$abcdef\ge C\cdot V^2$$

The trivial case $a=b=c=d=e=f$ gives $72\ge C$, but this is all I obtained.

This question is a 3D-version of inequality involving the product of lengths of edges of quadrilateral

The method used in RobertZ's solution seems not work here, since both sides would tend to 0.

• A better analogy in the planar case would be that of a triangle. The corresponding inequality should be $$abc\ge C\cdot A^{3/2}$$ and in that case I think one can prove $C=8/\sqrt{27}$. – Aretino Jul 25 '17 at 16:48
• Thanks for your comment. Yes, it's true. This inequality is equivalent to $\frac{1}{2}R\ge\frac{\sqrt{S\sqrt{3}}}{3}$ which is known and can be foung e.g. on imomath.com/index.php?options=604&lmm=0 (I wrote $S$ instead of $A$, because this letter is used on the page I gave link to) – tong_nor Jul 25 '17 at 17:38
• By analogy with that case I think $C$ attains its bound when all edges are equal, i.e. $C=72$. – Aretino Jul 25 '17 at 17:44

According to a chinese book Guide of Distance Geometric Analysis (距离几何分析导引) on my bookshelf, there is something called Veljan-Korchmáros inequality:

Let $v_1, v_2, \ldots, v_{n+1} \in \mathbb{E}^n$ be the vertices of a $n$-simplex. Let $a_{ij} = |v_i - v_j|$ be the edge lengths and $V$ be the volume of the simplex. We have $$\left(\prod_{1\le i < j \le n+1} a_{ij}\right)^{\frac{2}{n+1}} \ge n!\sqrt{\frac{2^n}{n+1}} V$$ and the equality is true when and only when the simplex is regular.

For $n = 3$, this implies $C = 72$.

I can't find an english reference of this inequality online. Following is the reference quoted in above book.

• Korchmáros G. Una limitazione per il volume di un simplesso $n$-dimensionale avente spigoli di date lunghezze. Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur., 1974, 56(6): 876-879.
• I corrected the reference, and found here a proof of the theorem. – Aretino Jul 25 '17 at 19:11
• @Aretino thanks for the fix, I just copied what is cited on the book which seems to insert spaces at random places. – achille hui Jul 25 '17 at 19:47

Let $a=b=c=d=e=1$ and $f=x$.

Hence, $V^2=\frac{x^2(3-x^2)}{144}$.

Thus, $$x\geq\frac{Cx^2(3-x^2)}{144}$$ or $$C\leq\frac{144}{x(3-x^2)}$$ and by AM-GM $$\frac{144}{3x-x^3}=\frac{144}{2-(2-3x+x^3)}\geq\frac{144}{2}=72$$

Thus, $C\leq72$, which is you know alredy.