= Solution
Use the following normalization for <Fourier analysis on a finite group>. Choose one <unitary irreducible representation> $\rho:G\to U(d_\rho)$ from each equivalence class, including the <trivial representation>. For a scalar function $f:G\to\mathbb C$, put
$$
\widehat f(\rho)=\mathbb E_x f(x)\rho(x),
\qquad
\langle f,g\rangle=\mathbb E_x\overline{f(x)}g(x).
$$
This convention uses $\rho(x)$, rather than $\rho(x)^*$, in the <Fourier transform on a finite group>; it makes the <normalized convolution on a finite group> preserve multiplication order.
The needed <representation theory> consists of <Maschke's theorem> and <unitarization of a finite-group representation>, together with the <Schur orthogonality relations>:
$$
\mathbb E_x\rho(x)_{ij}\overline{\sigma(x)_{kl}}
=\begin{cases}d_\rho^{-1}\delta_{ik}\delta_{jl},&\rho=\sigma,\\0,&\rho\ne\sigma.\end{cases}
$$
The <regular representation> contains $d_\rho$ copies of each $\rho$, so $\sum_\rho d_\rho^2=|G|$. Thus the scaled <matrix coefficients> $\sqrt{d_\rho}\rho(x)_{ij}$, and also their <complex conjugates>, form an <orthonormal basis> of all scalar functions on $G$. These facts imply <Fourier inversion on a finite group> and the <Parseval identity on a finite group> in the forms
$$
f(x)=\sum_\rho d_\rho\operatorname{tr}\bigl(\widehat f(\rho)\rho(x)^*\bigr),
\qquad
\langle f,g\rangle
=\sum_\rho d_\rho\operatorname{tr}\bigl(\widehat f(\rho)^*\widehat g(\rho)\bigr),
$$
and hence
$$
\|f\|_2^2=\sum_\rho d_\rho\|\widehat f(\rho)\|_{\mathrm{HS}}^2.
$$
In particular, the transform is an isomorphism onto the direct sum of the <matrix algebras> $M_{d_\rho}(\mathbb C)$, with the displayed weighted <Hilbert-Schmidt inner product>.
Define the <normalized convolution on a finite group> by
$$
(f*g)(x)=\mathbb E_yf(y)g(y^{-1}x).
$$
Substituting $x=yz$ and using the <group representation> identity yields the <convolution theorem on a finite group>
$$
\widehat{f*g}(\rho)=\widehat f(\rho)\widehat g(\rho).
$$
Unlike <normalized convolution on a finite group> on an <abelian group>, this product need not commute. If $\widetilde f(x)=\overline{f(x^{-1})}$, then $\widehat{\widetilde f}(\rho)=\widehat f(\rho)^*$. For <left translation of a group function> and <right translation of a group function> $L_af(x)=f(a^{-1}x)$ and $R_af(x)=f(xa)$,
$$
\widehat{L_af}(\rho)=\rho(a)\widehat f(\rho),
\qquad
\widehat{R_af}(\rho)=\widehat f(\rho)\rho(a)^*.
$$
For an <abelian group>, every <irreducible representation> is one-dimensional; this reduces to <Fourier analysis on a finite abelian group> with characters relabelled by their inverses. These formulas establish the basic scalar theory, with all normalizations and multiplication orders fixed.
Now suppose every nontrivial <irreducible representation> has $d_\rho\geq m$. If $f$ is a <mean-zero function>, its component at the <trivial representation> is zero. The <Parseval identity on a finite group> gives, for each other $\rho$,
$$
\|\widehat f(\rho)\|_{\mathrm{op}}
\leq\|\widehat f(\rho)\|_{\mathrm{HS}}
\leq\frac{\|f\|_2}{\sqrt{d_\rho}}
\leq\frac{\|f\|_2}{\sqrt m}.
$$
Using the <convolution theorem on a finite group>, the <Hilbert-Schmidt norm> inequality $\|AB\|_{\mathrm{HS}}\leq\|A\|_{\mathrm{op}}\|B\|_{\mathrm{HS}}$, and the <Parseval identity on a finite group> once more gives the <product mixing in a quasirandom group> estimate
$$
\boxed{\|f*g\|_2\leq m^{-1/2}\|f\|_2\|g\|_2\quad\text{when }\mathbb E f=0.}
$$
Write $a,b,c$ for the <subset density> values of $A,B,C$, respectively, and let $f=1_A-a$, $g=1_B-b$ be <balanced subset indicators>. Since both are <mean-zero functions>, $1_A*1_B=ab+f*g$. Their squared <norms> are $a(1-a)$ and $b(1-b)$. Also $\mathbb E(f*g)=0$, so the <Cauchy-Schwarz inequality> yields
$$
\begin{aligned}
\left|\mathbb E_z(1_A*1_B)(z)1_C(z)-abc\right|
&=\left|\mathbb E_z(f*g)(z)(1_C(z)-c)\right|\\
&\leq\sqrt{\frac{a(1-a)b(1-b)c(1-c)}m}
\leq\sqrt{\frac{abc}{m}}.
\end{aligned}
$$
If $abc>1/m$, the final bound is strictly smaller than $abc$. Thus the normalized number of pairs $(x,y)\in A\times B$ with $xy\in C$ is positive. Equivalently,
$$
\boxed{|A||B||C|>|G|^3/m\ \Longrightarrow\ AB\cap C\ne\varnothing.}
$$
This is the desired conclusion for a <quasirandom group>; the strict inequality ensures positivity rather than merely a nonnegative lower bound.
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