Cauchy–Schwarz inequality So I have a doubt regarding of the way I proved something and I am not sure it is good.
Let $V$ be an inner product space over $\Bbb C$ (the complex field).
Let $y \in V$ and let $x = \lambda y$, where $\lambda$ is a real number.
Then I have to prove that:
$$|(x,y)|= \sqrt{\langle x,x\rangle}\sqrt{\langle y,y\rangle}.$$
I know that (according to the inner product axioms):
$$\begin{align}
\sqrt{\langle x,x\rangle}\sqrt{\langle y,y\rangle}&=\sqrt{\langle\lambda y,\lambda y\rangle}\sqrt{\langle y,y\rangle} \\
&= \sqrt{\lambda \langle y,\lambda y\rangle}\sqrt{\langle y,y\rangle} \\ 
&= \sqrt{\lambda^2 \langle y, y\rangle}\sqrt{\langle y,y\rangle} \\
&= \sqrt{( \lambda \langle y, y\rangle)^2} \\
&= |\lambda|\langle y, y\rangle \quad \text{and because of the positive axiom we get that} \\
&= |\lambda ( y, y)|= |(x, y)|
\end{align}$$
and that is it.
But then I thought to myself that
$$\lambda\langle y, y\rangle= \sqrt{( \lambda \langle y, y\rangle)^2}= |\lambda|\langle y, y\rangle,$$
and if $ \lambda <0$ we get that 
$y=0$, which clearly is not the only case
So where is my mistake?
Thanks in advance!!
 A: Your computations are correct;  I think there is an inconsistency in
$$\lambda \langle y,y\rangle=\sqrt{(\lambda\langle y,y\rangle)^2}=|\lambda|\langle y,y\rangle, $$
however. The first equality should be $|\lambda|\langle y,y\rangle=\sqrt{(\lambda\langle y,y\rangle)^2},$ for all $\lambda\in\mathbb R$.
In particular, if $\lambda<0$, then $-\lambda\langle y,y\rangle=\sqrt{(\lambda\langle y,y\rangle)^2}.$
Question: why don't you choose $\lambda\in \mathbb C$?
Edit
If $\lambda\in \mathbb C$ , then $\langle\lambda y,y\rangle=\lambda\langle y,y\rangle$ , while $\langle y,\lambda y\rangle=\bar{\lambda} \langle y,y\rangle $,  or viceversa, according to conventions (if $\lambda$ is real you return to the formulae above, as $\lambda=\bar{\lambda}$). In other words 
$$(\langle \lambda y,y\rangle)(\langle y,\lambda y\rangle)=|\lambda|^2 (\langle y,y\rangle)^2.$$  
Mutatis mutandis, all is used is that  vector spaces over $\mathbb C$ are endowed with sesquilinear inner products, not bilinear ones.
