inequality involving the product of lengths of edges of a tetrahedron

Question

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: $$abcdef\ge C\cdot V^2$$

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

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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.

Answer 1

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.

Answer 2

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.