How to prove Schwarz inequality for Hermitian forms? 
I'm trying to do something like the proof of the Schwarz inequality for inner product.
If $h(y,y)\neq 0$, then we can take $\alpha=-h(x,y)/h(y,y)$ and calculate $h(x+\alpha y,x+\alpha y)$ which is a nonnegative number. The desired conclusion follows of this calculation.
If $h(y,y)=0=h(x,y)$, then the inequality is trivial.
My question is: how do we know that the case $h(y,y)=0$ and $h(x,y)\neq 0$ is not possible? When we are working with an inner product, this case is not possible because, by definition, $\langle y,y\rangle=0\Rightarrow y=0\Rightarrow\langle x,y\rangle=0$. But we don't have this condition for hermitian forms.
Thanks.
 A: If $h(y,y) = 0$ but $h(x,y) \ne 0$, then taking $\alpha = - t h(x,y)$ with $t > 0$
we would have 
$h(x + \alpha y,  x + \alpha y) = h(x, x) - 2 t |h(x,y)|^2$,
which would be negative for sufficiently large $t$.
A: The following proof is taken from S. Lang, Linear algebra.
Let $\alpha = h(w,w)$ and $\beta = -h(v,w)$. Then, writing $\|x\|^2 = h(x,x)$ for simplicity,
$$
0 \leq h(\alpha v + \beta w,\alpha v + \beta w) = \alpha \bar{\alpha} h(v,v) + \beta \bar{\alpha} h(w,v) + \alpha \bar{\beta} h(v,w) + \beta \bar{\beta} h( w,w).
$$
Hence
$$
0 \leq \|w\|^4 \|v\|^2 - 2 \|w\|^2 h(v,w) \overline{h(v,w)} + \|w\|^2 h(v,w) \overline{h(v,w)}
$$
and finally
$$
\|w\|^2 |h(v,w)|^2 \leq \|w\|^4 \|v\|^2.
$$
If $w=0$, the conclusion is trivial; otherwise we divide by $\|w\|^2$ and take a square root.
A: $h(y,y)=yHy$, in which $H$ is a PD matrix.
So $h(y,y)=0$ if.f $y=0$
In this case, $h(x,y)=xHy=xH0=0$
A: Instead of considering $h(y,y)$, you should consider $h(x,y)\neq 0$, then put $\alpha=\frac{-h(x,x)}{h(y,x)}$ – it always works. For $h(x,y)=0$, since $h(x,x)\ge 0$ we get the inequality.
