Let $z ∈ ℝ^n$, $r > 0$, and $ε ∈ (0, 2]$. Prove that if $x, y ∈$ B̅$(z,r)$ such that $||x - y|| ≥ εr$, then $||z - {{x+y} \over 2}|| ≤ r \sqrt{1 - {{ε^2} \over 4}}$.
my take on this exercise (not much, but I tried):
Knowing that B̅$(z,r)$ is a closed ball, we can say:
$||x-z|| ≤ r$ and $||y-z|| ≤ r$
From the triangle inequality we also know:
$| ||x|| - ||y|| ≤ ||x-y||$
How to continue to get the conclusion?
Let $\alpha=z-x$, $\beta=z-y$ and $\gamma=y-x=\alpha-\beta$, where $$|\alpha|^2,|\beta|^2\leq r^2,\qquad |\gamma|^2\geq \varepsilon^2 r^2.$$ We only need to prove $$(\alpha+\beta,\alpha+\beta)\leq r^2(4-\varepsilon^2).$$ Note that $$\begin{aligned} (\alpha+\beta,\alpha+\beta)&=|\alpha|^2+|\beta|^2+2(\alpha,\beta),\\ \varepsilon^2 r^2\leq |\gamma|^2=(\alpha-\beta,\alpha-\beta)&=|\alpha|^2+|\beta|^2-2(\alpha,\beta). \end{aligned}$$ It follows that $$\begin{aligned} |\alpha+\beta|^2&\leq 2(|\alpha|^2+|\beta|^2)-\varepsilon^2 r^2\\ &\leq (4-\varepsilon^2)r^2. \end{aligned}$$
