Verify by direct calculation that for any elements in an inner product space, $$\|z-x\|^2+\|z-y\|^2=\frac 12\|x-y\|^2+2\|z-\frac12(x+y)\|^2$$
How can I derive this identity without using the Parallelogram law? I am trying to do direct calculation as the problem is asking. I think it's easier to start on the right side of the equation.
All I have so far, in terms of using definitions of inner products: $$\|z-x\|^2+\|z-y\|^2=\langle z-x,z-x\rangle+\langle z-y,z-y \rangle$$ $$\frac 12\|x-y\|^2+2\|z-\frac12(x+y)\|^2=\frac12 \langle x-y,x-y \rangle+2 \langle z-\frac12(x+y),z-\frac 12(x+y)\rangle$$
If the vector space is over $\mathbb{C}$, we have $\|x+y\|^2=\|x\|^2+\|y\|^2+2\,\mathrm{Re}\langle x,y\rangle$. Of course, this holds in a vector space over $\mathbb{R}$, but there, $\mathrm{Re}\langle x,y\rangle=\langle x,y\rangle$.
$$ \begin{align} \|z-x\|^2+\|z-y\|^2 &=\langle z-x,z-x\rangle+\langle z-y,z-y\rangle\\[8pt] &=\|z\|^2+\|x\|^2-2\,\mathrm{Re}\langle z,x\rangle+\|z\|^2+\|y\|^2-2\,\mathrm{Re}\langle z,y\rangle\\[4pt] &=\frac12\left(\|x-y\|^2+\|x+y\|^2\right)+2\|z\|^2-2\,\mathrm{Re}\langle z,x+y\rangle\\ &=\frac12\|x-y\|^2+\frac12\left(\|x+y\|^2-4\,\mathrm{Re}\langle z,x+y\rangle+4\|z\|^2\right)\\ &=\frac12\|x-y\|^2+\frac12\|x+y-2z\|^2\\ &=\frac12\|x-y\|^2+2\left\|\frac{x+y}2-z\right\|^2 \end{align} $$