Proving an identity regarding character of irreducible representation 
Let $\frak{X}$ be an irreducible representation of a finite group $G$ affording the character $\chi$. Prove that for every $x,y\in G$: $$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z\in G}\chi(yzxz^{-1}).$$

My attempt: I know that I can replace $zxz^{-1}$ with $w$, and then the sum will be only on the conjugacy class of $x$, we'll denote it by $[x]$, but I don't know how many times every summand will appear after the change of variables, and to be honest I don't know how to take it from here.
Any help would be appreciated.
 A: 
Theorem Let $\chi \in Irr(G)$, and $x,y \in G$, then
$$\chi(x)\chi(y)=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi(xy^z)$$

Before proving this theorem we need to set notation and an observation. Write $\mathfrak{X}$ for any representation affording the character $\chi$. Let $x \in G$, then the $\color{blue}{conjugacy \ class}$ of $x$ in $G$ is denoted by $\color{blue}{K_x}$ and the $\color{darkgreen}{sum \ of \ its \ elements}$  as central element of the group algebra $\mathbb{C}[G]$ is denoted by $\color{darkgreen}{\hat{K}_x}$. As is well-known, the Schur Lemma implies the formula
$$\mathfrak{X}(\hat{K}_x)=\omega_{\chi}(\hat{K}_x)I$$
where $I$ is the identity matrix of dimension $\chi(1)$, and taking traces yields
$$\omega_{\chi}(\hat{K}_x)=\frac{\chi(x)|K_x|}{\chi(1)}$$
Observe that if we let $y$ run over $G$, and look at $x^y$, each element of $K_x$ appears $|C_G(x)|$ times. Hence we have the useful equality in $\mathbb{C}[G]$:

$$|C_G(x)|\hat{K}_x=\sum_{y \in G} x^y$$

Now let us proceed proving the theorem. Working in $\mathbb{C}[G]$ and applying the previous formula, we have
$$\sum_{z \in G}xy^z=\sum_{z \in G}xz^{-1}yz=x\sum_{z \in G}y^z=x|C_G(y)|\hat{K}_y$$
This gives
$$\mathfrak{X}(\sum_{z \in G}xy^z)=\sum_{z \in G}\mathfrak{X}(xy^z)=\mathfrak{X}(x)|C_G(y)|\omega_{\chi}(\hat{K}_y)I$$
Taking traces at both sides in the formula above gives
$$\sum_{z \in G}\chi(xy^z)=\chi(x)|C_G(y)|\omega_{\chi}(\hat{K}_y)=\chi(x)|C_G(y)|\frac{\chi(y)|\hat{K}_y|}{\chi(1)}=\chi(x)\chi(y)\frac{|G|}{\chi(1)}$$
which proves the theorem.$\square$
Corollary Let $\chi \in Irr(G)$. Then

$$\sum_{x,y \in G}\chi([x,y])=\frac{|G|^2}{\chi(1)}.$$

Proof In the Theorem, put $y=x^{-1}$. Then the formula yields
$$|\chi(x)|^2=\chi(x)\chi(x^{-1})=\frac{\chi(1)}{|G|}\sum_{z \in G}\chi([x^{-1},z]).$$
Since $\chi$ is irreducible we have $\sum_{x^{-1} \in G}|\chi(x^{-1})|^2=\sum_{x \in G}|\chi(x)|^2=|G|$, so summing over all $x \in G$ in the formula above gives the desired result.$\square$

Corollary Let $\chi \in Irr(G)$, and $x,y \in G$, then
$$\chi(x)\chi(y)=\frac{\chi(1)}{|G|^2}\sum_{g,h \in G}\chi(x^gy^h)$$

Proof The theorem tells us that
$$\sum_{g,h \in G}\chi(x^gy^h)=\sum_{g\in G}(\sum_{h \in G}\chi(x^gy^h))=\sum_{g\in G}\frac{|G|}{\chi(1)}\chi(x^g)\chi(y)=\sum_{g\in G}\frac{|G|}{\chi(1)}\chi(x)\chi(y)=\frac{|G|^2}{\chi(1)}\chi(x)\chi(y).$$
