In a space with inner product show that $\|x+y\|^2+\| x-y\|^2=2\left \| x \right \|^2+2\left \| y \right \|^2$
after a lot of tries I ended up with : $\begin{Bmatrix} \langle x,x\rangle +\langle -y,x\rangle=\langle x-y,x\rangle \\ \langle -y,-y\rangle+\langle -y,x\rangle=\langle -y,x-y\rangle\\ \langle x,x\rangle+\langle y,x\rangle=\langle x+y,x\rangle\\ \langle y,y\rangle+\langle y,x\rangle=\langle x+y,y\rangle \end{Bmatrix}$
If we add them all I think we get what we want, if we consider the fact that $\langle -y,-y\rangle=\langle y,y\rangle$ is this correct?
Does this equation have a geometric meaning? I think the inner product measures the angle and the length (not sure about that) but I can't warp my head around what this exactly says.
Just write $$ \|x+y\|^2 = \langle x+y, x+y\rangle, \quad \|x-y\|^2 = \langle x-y, x-y\rangle, $$ and use the properties of the inner product.
Note that the inner product is bilinear. In this case, $$ \|x+y\|^2=\langle x+y, x+y \rangle = \langle x,x\rangle + \langle x,y \rangle +\langle y,x\rangle + \langle y,y\rangle = \|x\|^2+\|y\|^2 + 2 \langle x,y\rangle $$
$$ \|x-y\|^2=\langle x-y, x-y \rangle = \langle x,x\rangle - \langle x,y \rangle -\langle y,x\rangle + \langle y,y\rangle = \|x\|^2+\|y\|^2 - 2 \langle x,y\rangle $$
Although I'm assuming symmetry in the last step, the main result holds regardless of that.