Apollonius’ Identity inner product space

Question

$||z-x||^2+||z-y||^2=\frac{1}{2}||x-y||^2+2||z-\frac{x+y}{2}||^2$

I proved it by expanding both sides and i found both sides are equal. Are there any easy way to prove it?

Answer 1

In a Hilbert space you can use the Parallelogram law :

$$\|a+b\|^2 + \|a-b\|^2 = 2(\|a\|^2 + \|b\|^2)$$

So :

$$\begin{align} &\|z - \frac{x+y}{2} \|^2\\ &= \|\frac{z}{2} - \frac{x}{2} + \frac{z}{2} - \frac{y}{2}\|^2\\ &= 2(\|\frac{z}{2} - \frac{x}{2}\|^2 + \|\frac{z}{2} - \frac{y}{2}\|^2) - \|\frac{y}{2} + \frac{x}{2}\|^2\\ &= \frac12(\|z - x\|^2 + \|z - y\|^2 - \frac{1}{4}\|y - x\|^2\\ &\square\end{align}$$

Answer 2

Consider the Parallelogram Rule: $2\bigg(\|X\|^2+\|Y\|^2 \bigg)=\|X+Y\|^2+\|X-Y\|^2$

Now apply the transformation $X=z-x\ \&\ Y=z-y$ and result holds.