I am reading "Strong uniform times and finite random walks" by Aldous and Diaconis, and I got stuck in a inequality, and would appreciate some help.
Consider a finite group $G$, $\hat{G}$ its space of irreducible representations, a distribution $Q$ on $G$, which is constant on conjugacy classes, and $\tilde{Q}(\rho) = \displaystyle \frac{1}{d_{\rho}}\sum_{g\in G} Q(g) \chi_{\rho}(g)$, for every $\rho \in \hat{G}$, where $\chi_{\rho}(g):= trace(\rho(g))$, is the character associated with $\rho$, and $d_{\rho}$ is the degree of the representation $\rho$. I would like to see why
$$1-|G| \leq \displaystyle \sum_{\rho \in \hat{G}}d_{\rho}|\chi_{\rho}(g)\tilde{Q}(\rho)|.$$
What I've tried so far:
If we start from $1-|G|=|G|(\dfrac{1}{|G|}-Q(g)) = |G|(U(g)-Q(g))$ where $U$ is the uniform distribution on $G$. We could apply upper bound lemma to have
$$|G|(U(g)-Q(g)) \leq |G|||U-Q|| \leq |G| \displaystyle \sqrt(\frac{1}{4}\sum_{\rho}d_{\rho}||\hat{Q}(\rho)||^2,$$
where $\hat{Q}$ is the Fourier transform of $Q$, defined as $\hat{Q}(\rho)= \displaystyle \frac{1}{|G|}\sum_{g\in G} Q(g)\rho(g)$. But anything from here doesn't fit to the inequality. The same happens if you try to start from the right-hand side of the inequality.
Any comments or suggestions are welcome.