Cauchy-Schwarz Inequality without using $\langle a x,y\rangle=a\langle x,y\rangle$ Let $V$ be a vector space and define a function $\langle .,.\rangle:V\times V\to\mathbb{C}$ such that 

$$\begin{align}
& \langle x,y\rangle=\overline{\langle y,x\rangle }\,\,\,\forall x,y\in V\\
& \langle x+y,z\rangle=\langle x,z\rangle+\langle y,z\rangle
\,\,\,\,\,\forall x,y,z\in V\\ 
& \langle x,x\rangle\ge0\,\,\,\,\,\forall x\in V\,\,\,\text{and equality holds iff}\,\,\,\ x=0\\ 
\end{align}$$

Does Cauchy-Schwarz inequality $$\color{Green}{|\langle x,y\rangle|^2\le\langle x,x\rangle\langle y,y\rangle}$$still vallied?
  (Without the condition $\langle a x,y\rangle=a\langle x,y\rangle,\,\,\,\forall a\in\mathbb{C},\,\,\,\forall x,y\in V.$)   
If the answer is "NO", can we prove it?
 A: The answer is no in general.
For example with $V = \mathbb{C}$, it's not too hard to see that "inner product" with
$$
\langle x_1 + iy_1, x_2 + iy_2 \rangle = x_1x_2+y_1y_2+2i(x_1y_2-y_1x_2)
$$
satifies all the other conditions, but we have $\langle 1, 1 \rangle = 1 = \langle i, i \rangle$, and $\langle 1, i \rangle = 2i$, so the Cauchy-Schwarz doesn't hold for $x,y = 1, i$. And you could certainly cook up similar other examples.
Having $\mathbb{C}$ as the scalar field is crucial here though. With $\mathbb{R}$ as the field, we get the inequality: since for any "inner product" we have $\langle 2x, y \rangle = \langle x, y \rangle + \langle x, y \rangle = 2 \langle x, y \rangle$ by induction
$\langle nx, y \rangle = n\langle x, y \rangle$ for any natural number. It's quite easy to extend to integers, and finally with 
$$
p\langle x, y \rangle = \langle px, y \rangle = \left\langle q\frac{p}{q}x, y \right\rangle = q\left\langle \frac{p}{q}x, y \right\rangle
$$
to rationals. Now a standard proof works out: for any rational $t$ we have
$$
0 \leq \langle x + ty, x + ty \rangle = \langle x, x \rangle + 2 t\Re{\langle x, y \rangle} + t^2\langle y, y \rangle.
$$
Since the inequality holds for any rational $t$, it does for any real $t$, so for discriminant $D$ we get
$$
0 \geq D = (2\Re{\langle x, y \rangle})^2-4\langle x, x \rangle\langle y, y \rangle,
$$
so finally
$$
\langle x, x \rangle\langle y, y \rangle \geq (\Re{\langle x, y \rangle})^2 = |\langle x, y \rangle|^2
$$
