Let $G$ be a finite group and $\Lambda = (\lambda\_{i,j})$ its character table with $\lambda\_{i,1}$ the degree of the ith character. 


Consider the following combinatorial property of $\Lambda$: for all triple $(j,k,\ell)$ $$\sum\_i \frac{\lambda\_{i,j}\lambda\_{i,k}\lambda\_{i,\ell}}{\lambda\_{i,1}} \ge 0.$$
It is a consequence of a more general result involving subfactor planar algebra and fusion category (see [here](https://arxiv.org/abs/1910.12059v1) Corollary 7.5, see also [this answer](https://mathoverflow.net/a/344806/34538)).


*Question*: Is this combinatorial property already known to finite group theorists?  

If yes: What is a reference?  

If no: Is there a group theoretical *elementary* proof?  

*In any case*: Are there other properties of the same kind?




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To avoid any misunderstanding, let us see one example. Take $G=A\_5$, its character table is:  

$$\left[ \begin{matrix}
1&1&1&1&1 \\
3&-1&0&\frac{1+\sqrt{5}}{2}&\frac{1-\sqrt{5}}{2} \\
3&-1&0&\frac{1-\sqrt{5}}{2}&\frac{1+\sqrt{5}}{2} \\
4&0&1&-1&-1 \\
5&1&-1&0&0
\end{matrix} \right] $$
Take for example $(j,k,\ell) = (2,4,5)$, then $\sum\_i \frac{\lambda\_{i,j}\lambda\_{i,k}\lambda\_{i,\ell}}{\lambda\_{i,1}} = \frac{5}{3} \ge 0$.



A new combinatorial property for the character table of a finite group?
By standard manipulations with the group algebra, your sum has a combinatorial/probabilistic interpretation that makes its nonnegativity clear.


The element $ \frac{1}{|G|} \sum\_{ g\in G} [g hg^{-1} ]$ in the group algebra is conjugacy invariant, and so acts by scalars on each irreducible representation. Because its trace on a representation with the character $\chi$ is $ \frac{1}{|G|} \sum\_{ g\in G} \chi( g hg^{-1} ) = \frac{1}{|G|} \sum\_{ g\in G} \chi( h )= \chi(h)$, its unique eigenvalue must be $\frac{\chi(h)}{\chi(1)}$. Hence for $h\_1,h\_2,h\_3$ three elements of the group, 


$$ \left( \frac{1}{|G|} \sum\_{ g\in G} [g h\_1g^{-1} ]\right) \left( \frac{1}{|G|} \sum\_{ g\in G} [g h\_2g^{-1} ]\right) \left( \frac{1}{|G|} \sum\_{ g\in G} [g h\_3g^{-1} ]\right) $$


acts on this representation with eigenvalue $\frac{ \chi(h\_1) \chi(h\_2) \chi(h\_3)}{\chi(1)^3}$. 


Now the group algebra, as a module over itself, is the sum over irreducible characters $\chi$ of $\chi(1) $ copies of the representation with character $\chi$. Hence the trace of this element on the group algebra is $$\sum\_{\chi} \chi(1) \cdot \chi(1) \cdot \frac{ \chi(h\_1) \chi(h\_2) \chi(h\_3)}{\chi(1)^3}= \sum\_{\chi} \frac{ \chi(h\_1) \chi(h\_2) \chi(h\_3)}{\chi(1)}.$$


On the other hand, the trace of an element of the group algebra on itself is the order of the group times the coefficient of $[1]$. The coefficient of $[1]$ in this particular element is $\frac{1}{ |G|^3}$ times the number of $g\_1,g\_2,g\_3$ such that $g\_1 h\_1 g\_1^{-1} g\_2 h\_2 g\_2^{-1} g\_3 h\_3 g\_3^{-1} =1$. This gives the combinatorial interpretation


$$\sum\_{\chi} \frac{ \chi(h\_1) \chi(h\_2) \chi(h\_3)}{\chi(1)} = \frac{1}{ |G|^2} \left| \{ g\_1,g\_2,g\_3 \in G \mid g\_1 h\_1 g\_1^{-1} g\_2 h\_2 g\_2^{-1} g\_3 h\_3 g\_3^{-1} =1 \}\right|$$


from which non-negativity is clear.


I would guess this is probably in the group theory literature somewhere but I wouldn't know where.


