Proof of Cauchy-Schwarz and the parallelogram equality Let $ V $ be a normed vector space over $\mathbb{R}$ with a norm $ \lVert\cdot\rVert$ that satisfies the parallelogram equality.
We define the inner product :
$$\langle x, y \rangle = \frac{1}{4} \left( \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right).$$
and I need to prove that Cauchy-Schwarz inequality holds with this inner product.
Here is my attempt: 
$$|\langle x, y \rangle | = \frac{1}{4} \left| \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right| \leq \frac{1}{4} \left( \Vert x + y \Vert^2 + \Vert x - y \Vert^2 \right) = \frac{1}{4} \left( 2 \left( \Vert x  \Vert^2 + \Vert y \Vert^2 \right) \right) = \frac{1}{2}\left( \Vert x  \Vert^2 + \Vert y \Vert^2  \right)  $$

There is no error in you calculations, but they wouldn't give you the result you're hoping for;
The inequality $$\frac{1}{4} \left| \Vert x + y \Vert^2 - \Vert x - y \Vert^2\right| \leq \frac{1}{4} \left( \Vert x + y \Vert^2 + \Vert x - y \Vert^2 \right) $$
is too weak for this manner.
Hint: Try separating the proof to $\langle x, y \rangle > 0 $ and $\langle x, y \rangle < 0 $, then try using the parallelogram equality + the norm's triangle inequality for each case.
