Let $\langle \cdot, \cdot \rangle$ be a real positive definite hermitian form in Hilbert space $H$. Prove that if $\langle x,x\rangle =0$ then $x=0$

Question

I think it may be connected to the Schwarz inequality.

Because if $\langle x, x \rangle = 0$ implies the Schwarz inequality, then $\langle x, y \rangle = 0$ too for $x \neq y$.

And for some reason, because $\langle x, y \rangle = 0$ for $x \neq y$ when $\langle x, x \rangle = 0$, this means that $x = 0$.

It's clear that $\langle x, x \rangle = 0$ for $x = 0$, but what we have to prove is that $\langle x , x \rangle = 0 \ \Rightarrow \ x = 0$. So theoretically, there might be other non-zero $x$ which will fulfill $\langle x, x \rangle = 0$.