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 Corollary 7.5, see also this answer).
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?
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 $\sum_i \frac{\lambda_{i,j}\lambda_{i,k}\lambda_{i,\ell}}{\lambda_{i,1}} = \frac{5}{3} \ge 0$.
Best Answer
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.