Let $χ$ be an irreducible character of finite group $G$. Prove that $$χ(e)^2\leq\big[G: Z(χ)\big]\,,$$ and the equality holds iff $χ(g) = 0$ for all $g ∈ G\backslash Z(χ)$.
Any help would be appreciated. Note here that $$[G: Z(χ)] = |G/Z(χ)| = \frac{|G|}{|Z(χ)|}$$ where $G\backslash Z(χ)$ is the set difference and $G/Z(χ)$ is the set of cosets.
Since $\chi$ is an irreducible character, $$ \frac{1}{|G|}\,\sum_{g\in G}\,\big|\chi(g)\big|^2=\langle \chi,\chi\rangle=1\,.$$ Therefore, $$\sum_{g\in Z(\chi)}\,\big|\chi(g)\big|^2\leq \sum_{g\in G}\,\big|\chi(g)\big|^2=|G|\,.$$ However, as $\big|\chi(g)\big|=\chi(e)$ for all $g\in G$ (presumably, $e$ is the identity element of $G$), we get $$\big|Z(\chi)\big|\,\big|\chi(e)\big|^2=\sum_{g\in Z(\chi)}\,\big|\chi(g)\big|^2\leq |G|\,.$$ Therefore, $$\big(\chi(e)\big)^2=\big|\chi(e)\big|^2\leq \frac{|G|}{\big|Z(\chi)\big|}=\big[G:Z(\chi)\big]\,.$$ Clearly, the equality holds if and only if $$\sum_{g\in Z(\chi)}\,\big|\chi(g)\big|^2= \sum_{g\in G}\,\big|\chi(g)\big|^2\,,$$ which means $\chi(g)=0$ for every $g\in G\setminus Z(\chi)$.