Question: > 
> Let $V$ be a vector space over a field $\mathbb K$ and $f$ be a nonnegative Hermitian form on $V$. Then, $\forall x,y \in V$:
> 
> 
> $$|f(x,y)|^2 \le f(x,x)f(y,y)$$
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> 
> 

Here's one proof:

For an arbitrary $\alpha \in \mathbb K$, compute: $f(x + \alpha y, x + \alpha y)$ and $f(\alpha x + y, \alpha x + y)$ and note that both of these are nonnegative quantities. Then, separate the cases $f(y,y) \neq 0$ and $f(y,y) = 0$ (where the latter separated into the cases $f(x,x) \neq 0$ and $f(x,x) = 0$), each time with a suitable choice of $\alpha$ to obtain the result.

My question is:

> 
> Is there another proof, perhaps a nicer one or one which uses a sneaky approach?
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> 
> 

Thank you.

Answer: 
*I'm modifiyng Bernard's approach for nonnegative Hermitian form.*  

Let $t\in \mathbb R$ and $\mathbb K \in \{ \mathbb R,\mathbb C \}$ , then $$0\leq f(tx+y,tx+y)=t^2f(x,x)+t(f(x,y)+\overline{f(x,y)})+h(y,y)$$
Since $z+\overline z = 2Re(z)$, it follows 
$$0\leq f(tx+y,tx+y)=t^2f(x,x)+2tRe(f(x,y))+h(y,y).$$
It's discriminant must satisfy $Re(f(x,y))^2-f(x,x)f(y,y)\leq0$,so we proved
$$ Re(f(x,y))^2\leq f(x,x)f(y,y) \quad \forall x,y\in \mathbb C$$
Let $\phi:=Arg(f(x,y))$, now substitute $ye^{i\phi}$ for $y$ and note that $Re(f(x,ye^{i\phi})=\underbrace{e^{i\phi}\overline{e^{i\phi}}}\_{1}|f(x,y)|$.
You have alternate (and simpler) proofs on [Cauchy-Schwarz wikipedia](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality#Proof)