Show that a positive operator on a complex Hilbert space is self-adjoint

You should apply the polarization identity in the form

$$4(Ax,y) = (A(x+y),x+y) - (A(x-y),x-y) -i(A(x+iy),x+iy) + i(A(x-iy),x-iy).$$

Since you already know $(Az,z) = (z,Az)$ for all $z \in \mathcal{H}$, it is not difficult to deduce $A^\ast = A$ from that.

Another solution is this one:

$T$ is self-adjoint iff $\langle Tx,x\rangle \in \mathbb{R}$ for every $x\in \mathcal{H}$.

Let $x\in \mathcal{H}$. Then $0=\langle Tx,x \rangle-\langle T^{*}x,x \rangle = \langle Tx,x \rangle-\overline{\langle Tx,x \rangle}=2i\operatorname{Im}\langle Tx,x\rangle \iff \langle Tx,x \rangle \in \mathbb{R}$. Thus $T=T^{*} \iff \langle Tx,x\rangle \in \mathbb{R}$ for every $x\in \mathcal{H}$.

Because $T$ is positive we have that $\langle Tx,x \rangle \in \mathbb{R}$.