Let $G$ be a finite (not necessarily abelian) group and let $S$ be a symmetric generating set of $G$, i.e. if $s\in S$ then $s^{-1} \in S$.
Let $\chi$ be an irreducible character of $G$. I have managed to show that $ \sum_{g_1,g_2,g_3,g_4 \in S}\chi(g_1 g_2g_3g_4) \geq 0 $.
How? Rewrite the LHS as $\chi((\sum_{g\in S}g)^4)$, and use the star operator (which takes $\sum_{i}a_ig_i$ to $\sum_{i}a_i^*g_i^{-1}$), which preserves $\sum_{g\in S}g$ (since $S$ is a symmetric generating set). Then, conclude that the LHS is the same as $\chi(((\sum_{g\in S}g)(\sum_{g\in S}g)^*)^2)$. Finally, if $A$ is the corresponding matrix to $\sum_{g\in S}g$, the above is just $Tr((AA^*)^2)$, which is clearly non-negative.
I would like to generalize this result to more than a single character. For example, I would like to show that $\sum_{g_1,g_2,g_3,g_4,g_5,g_6\in S}\chi_1(g_1g_2g_3g_4)\cdot \chi_2(g_3g_4g_5g_6) \geq 0$ for every two irreducible characters $\chi_1,\chi_2$.
What can we do to show that? I tried using the same technique but didn't manage to complete the calculation.
Proposed solution:
Rewrite the sum as $\sum_{g_3,g_4 \in S}\chi_1((\sum_{s\in S}s)^2 g_3g_4)\cdot \chi_2((\sum_{s\in S}s)^2 g_3g_4)$. Then remember that $\chi =\chi_1 \cdot \chi_2$ is a character of the tensor product of the correspoding representations for $\chi_1,\chi_2$, hence we can rewrite $\sum_{g_3,g_4 \in S}\chi((\sum_{s\in S}s)^2 g_3g_4) = \chi((\sum_{s\in S}s)^4)$ and this is non-negative due to the same argument in the single character case. Can you find any incorrect argument here?