How big is a tetrahedron? Let $T$ be a tetrahedron with volume $vol(T)$ and edge lengths $a,b,c,d,e,f$ and
let $sum(T) = a^3 + b^3 + ... + f^3$.  We wish to compare $vol(T)$ with $sum(T)$.  [ IMO (1961 #2 ) handles the case of triangles. ]
Either by approaching it directly or by expressing the volume as a determinant,
invoking Hadamard's bound, and applying the AM-GM inequality, one finds that
$vol(T) < sum(T)/36$.  However, this hardly seems best possible.
Question: What is the sharpest upper bound for $vol(T)$ in terms of $sum(T)$?
Extra: Generalize to $n$ dimensions.
 A: 
Generalize to $n$ dimensions.

As you wish. :-) But you don’t fool me: IMO 1961 #2, also known as Weitzenböck’s inequality, is related with squares of edge lengths, not cubes. :-) So, it seems the following.
Let $T\subset\Bbb R^n$ be an $n$-dimensional simplex, $p\ge 2$ be a real number, $\mathcal L(T)$ be the set of all ($1$-dimensional) edges of the simplex $T$ and 
$$s_p(T)=\sum_{l\in \mathcal L(T)}\|l\|^p.$$ By Power Mean Inequality, if $p<q$ then 
$$\left(\frac{s_p(T)}{n\choose 2}\right)^\frac 1p\le \left(\frac{s_q(T)}{n\choose 2}\right)^\frac 1q,$$
and for a regular simplex (which has all edges of equal length) all these inequalities become equalities. 
So if we show that $$vol(T)\le C(s_p(T))^\frac{n}{p}$$ for any simplex $T$, then we automatically obtain that 
$$vol(T)\le Cs_q(T)^\frac{n}{q}{n\choose 2}^\frac{n(q-p)}{pq}$$
for each $q\ge p$. 
Conversely, if we show that $$vol(T)=C(s_p(T))^\frac{n}{p}$$ for a regular simplex $T$ (which has all edges of equal length), then  we automatically obtain that 
$$vol(T)=Cs_q(T)^\frac{n}{q}{n\choose 2}^\frac{n(q-p)}{pq}$$
for each $q\ge p$. 
So, let’s show it for a function $s(T)=s_2(T)$. For this purpose first of all we fix an ($n$-dimensional) volume $V=vol(T)$ of the simplex $T=\text{conv}\{v_0,\dots, v_n\}\subset\Bbb R^{n}.$. Also $T$ can be considered as a point $(v_0,\dots, v_n)$ of the space $\Bbb R^{n\cdot (n+1)}$. Put $\mathcal T=\{T\in R^{n(n+1)}:vol(T)=V\}$. The set $\mathcal T$ is closed in $\Bbb R^{n(n+1)}$, because $V(T)$ is a continuous function of the coordinates of vertices of $T\in \Bbb R^{n(n+1)}$. Since the function $s(T)$ defined on the family $\mathcal T$ goes to infinity when a length of at least one of the edges of $T$ goes to infinity, the function $s(T)$ attains its minimum at some simplex $T\in\mathcal T$.
Let $v_i$ be a vertex of the simplex $T$. Fix all vertices of $T$ except for $v_i$ and move $v_i$ along a hyperplane $H_i$ spanned by the fixed vertices. This motion does not change the volume of $T$, but changes a sum $s(v_i)$ of the squares of lengths of edges of the simplex $T$  which are incident to $v_i$. Let 
$$f_i=\frac 1{n}\sum_{j\ne i} v_j$$
be the centroid of the ($(n-1)$ dimensional) face (considered as a set of its vertices) opposite to the vertex $v_i$.  I claim that 
$$s(v_i)=\sum_{j\ne i}\|v_j-f_i\|^2 +n\|v_i-f_i\|^2.$$
Indeed,
$$\sum_{j\ne i}\|v_j-f_i\|^2 +n\|v_i-f_i\|^2=$$
$$\sum_{j\ne i}(v_j-f_i, v_j-f_i) +n(v_i-f_i, v_i-f_i)=$$ 
$$\sum_{j\ne i}\left(v_j-\frac 1{n}\sum_{k\ne i}v_k, v_j-\frac 1{n}\sum_{k\ne i} v_k\right) +n\left(v_i-\frac 1{n}\sum_{k\ne i} v_k, v_i-\frac 1{n}\sum_{k\ne i} v_k\right)=$$
$$\sum_{j\ne i}\left[(v_j,v_j)- \frac 2{n}\left(v_j,\sum_{k\ne i} v_k\right)+ \frac 1{n^2}\left(\sum_{k\ne i}v_k, \sum_{k\ne i} v_k\right)\right]+$$ $$n(v_i,v_i)- 2\left(v_i,\sum_{k\ne i} v_k\right)+ \frac 1{n}\left(\sum_{k\ne i}v_k, \sum_{k\ne i} v_k\right)=$$
$$\sum_{j\ne i}(v_j,v_j)+n(v_i,v_i)- 2 \left(v_i,\sum_{j\ne i} v_j\right)=$$
$$\sum_{j\ne i}\left[(v_j,v_j)+(v_i,v_i)- 2(v_i,v_j)\right]=$$
$$\sum_{j\ne i}(v_j-v_i, v_j-v_i)=\sum_{j\ne i}\|v_j-v_i\|^2=s(v_i).$$
Therefore the minimum of $s(v_i)$ is attained iff the minimum of $\|v_i-f_i\|$ is attained iff $v_i-f_i$ is orthogonal to $H_i$. 
Then for the case of minimal $s(T)=\frac 12\sum s(v_i)$ we have that all heights of the simplex $T$ are concurrent in the centroid $O$ of $T$ (considered as a set of its vertices). Now fix the origin at the point $O$. Then $\sum v_i=0$ and each vector $v_i$ is orthogonal to the difference $v_j-v_k$ for all $j\ne i\ne k$. Therefore $(v_i, v_j)=(v_i,v_k)=a$.  Then $$(v_i,v_i)=(v_i,v_i-\sum v_j)=-na$$ and 
$$\|v_i-v_j\|^2=(v_i-v_j,v_i-v_j)=(v_i,v_i)-2(v_i,v_j)+(v_j,v_j)=-2na-2a=\text{const}.$$ 
So the simplex $T$ is regular.$\square$
A: If $a.b.c.d.e.f$ are the edges of a 4-hedron; and   
if $a$ opposes $d$, $b$ opposes $e$, and $c$ opposes $f$; and   
if $\{A,B,C,D,E,F\}=\{a^2,b^2,c^2,d^2,e^2,f^2\}$, respectively:    
    
then $Q=AD(-A+B+C-D+E+F)$
$+BE(A-B+C+D-E+F)$
$+CF(A+B-C+D+E-F)$
$-ABC-AEF-BDF-DEF$; and
volume $=\frac{\sqrt{Q}}{12}$.
