Suppose that $x,y,z\in \mathbb{R}^n$ such that $\|y\| \le 1, \|x-y\|\le 4,\|y-z\| \ge 5$ Show that $ |\|x-y\| ^2-\|x\|^2|< \|x\|^2+\|z\|^2$

Question

Suppose that $x,y,z$ are vectors from $\mathbb{R}^n$ such that $\|y\| \le 1, \|x-y\|\le 4,\|y-z\| \ge 5$. Show that

$$ |\|x-y\| ^2-\|x\|^2|< \|x\|^2+\|z\|^2$$

How to prove this?

Here I know that $|\|x\|^2 - \|y\|^2| \le \|x-y\|\|x+y\| $

So, $|\|x-y\| ^2-\|x\|^2|\le \|x-y+x\|\|x-y-x\|=\|y\|\|2x-y\|$

Answer 1

Note that $$\tag1 5\geq\|x\|\geq3. $$ Because if $\|x\|<3$ you get $\|x-y\|\leq\|x\|+\|y\|<3+1=4$. And $\|x\|\leq \|y\|+4=5$. Also, $$\tag2 \|z\|\geq4, $$ because otherwise we get $5\leq\|y-z\|\leq\|y\|+\|z\|<1+4=5$. Now \begin{align} \big||x-y\|^2-\|x\|^2\big| &=\|x\|^2+\|y\|^2-2\,\langle x,y\rangle-\|x\|^2\\[0.3cm] &=\|y\|^2-2\langle x,y\rangle\\[0.3cm] &\leq1+2\|x\|\\[0.3cm] &\leq11\leq 3^2+4^2\\[0.3cm] &\leq\|x\|^2+\|z\|^2. \end{align}