A conjugacy class of more than average size must have character value zero for some irreducible character. I found the following claim here and wondered how to go about proving it?
Let $C$ be a conjugacy class of $G$ of more than average size, so $\frac{\#{G}}{\#{C}} <$ Number of Conjugacy Classes.
Show that there is an irreducible character $\chi$ of $G$ over $\mathbb{C}$ such that $\chi = 0$ on $C$.
 A: Proposition Let $Cl_G(g)$ be a conjugacy class of the finite group $G$ with $|C_G(g)| \lt k(G)$, the number of conjugacy classes of $G$. Then $\chi(g)=0$ for some $\chi \in Irr(G)$.
Proof By the Second Orthogonality Relation we have $|C_G(g)|=\sum_{\chi \in Irr(G)}|\chi(g)|^2$, hence $1 \gt \frac{1}{k(G)}\sum_{\chi \in Irr(G)}|\chi(g)|^2$. Let $G$ have exponent $n$, and put $K=\mathbb{Q}(e^{\frac{2\pi i}{n}})$, then by a theorem of R. Brauer (see Isaacs, CTFG, (10.3) Theorem) every $\chi \in Irr(G)$ is afforded by a $K$-representation. $\mathcal{G}=Gal(K/\mathbb{Q})$ acts on $Irr(G)$ by $\chi^{\sigma}(g)=\chi(g)^{\sigma}$ for every $\sigma \in \mathcal{G}$. Let $\mathcal{I}_1, \cdots ,\mathcal{I}_t$ be orbits of $Irr(G)$ under the action of $\mathcal{G}$. The geometric-arithmetic mean formula yields
$$1 \gt \frac{1}{k(G)}\sum_{\chi \in Irr(G)}|\chi(g)|^2 \geq (\prod_{\chi \in Irr(G)}|\chi(g)|^2)^{\frac{1}{k(G)}}=(\prod_{j=1}^t\prod_{\chi \in \mathcal{I}_j}|\chi(g)|^2)^{\frac{1}{k(G)}}$$ and hence
$$1 \gt \prod_{j=1}^t\prod_{\chi \in \mathcal{I}_j}|\chi(g)|^2 \geq 0.$$
Since complex conjugation commutes with any element of $\mathcal{G}$, each of the factors $\alpha_i=\prod_{\chi \in \mathcal{I}_j}|\chi(g)|^2$ is fixed by $\mathcal{G}$ and is an algebraic integer, each $\alpha_i$ is a rational integer. This can only happen when at least one of the $\alpha_i$'s equals zero. Hence $\chi(g)=0$ for some $\chi \in Irr(G)$.
