Positive linear operator inequality

Question

I'm stuck proving the following inequality. Let $H$ autohermitian operator such that $$\langle u|H|u\rangle\geq0\qquad\forall |u\rangle$$ Proof that $$|\langle u|H|v\rangle|^2\leq\langle u|H|u\rangle\langle v|H|v\rangle$$ with equality $\langle u|H|u\rangle=0$ if $H|u\rangle =0$.

Can you help me with an idea please?

Answer 1

This is Cauchy-Schwarz for the positive form $$[x,y]=\langle x|H|y\rangle.$$ Just use or repeat the proof of Cauchy-Schwarz.