Let $z\in \mathbb{R}^n, r > 0$, and $\epsilon \in (0,2]$. Prove that if $x, y \in \overline{B}(z,r)$ such that $\|x-y\| \ge \epsilon r,$ then $ \|z-\frac{x+y}{2}\|\le r \sqrt{1-\frac{\epsilon ^ 2}{4}}$. So far I have:
$$\|z - \frac{x+y}{2}\| = \sqrt{\langle z-\frac{x+y}{2},z-\frac{x+y}{2} \rangle}$$
Then:
$$\langle z - \frac{x+y}{2}, z-\frac{x+y}{2}\rangle = \langle z - x, z-\frac{x+y}{2}\rangle + \langle \frac{x-y}{2}, z-\frac{x+y}{2}\rangle $$$$=\langle z-x, z- y \rangle + \langle z - x, \frac{y-x}{2}\rangle + \langle \frac{x-y}{2}, \frac{y-x}{2}\rangle + \langle \frac{x-y}{2}, z-y\rangle $$ $$=\langle z-x,z-y\rangle + \langle \frac{x-y}{2}, x-y\rangle-1/4\|x-y\|^2$$ $$=\langle z-x, z-y\rangle+1/2\|x-y\|^2-1/4\|x-y\|^2$$ $$\le\|z-x\|\|z-y\|-\frac{(\epsilon r)^2}{4}+1/2\|x-y\|^2$$ $$\le r^2-\frac{(\epsilon r)^2}{4}+1/2\|x-y\|^2=r^2(1-\frac{\epsilon^2}{4})+1/2\|x-y\|^2$$
I think I'm on the right track but I keep getting the $1/2\|x-y\|^2$ and I don't know what to do. Help?
By translating by $z,$ you can assume $z = \vec{0}$ for simplicity. Then,
$$\|x+y\|^2+\|x-y\|^2 = 2\|x\|^2+2\|y\|^2\leq4r^2\implies \|x+y\|^2\leq 4r^2-\epsilon^2 r^2$$
and you are done.