Is there a possibility to obtain a bound on the difference of squared norm $\left\lvert\Vert x \Vert^2 – \Vert y \Vert^2\right\rvert$ in terms of the norm of the difference $\Vert x – y \Vert$?
I have tried to adapt the usual proof of the reverse triangle inequality as follows.
$$
\Vert x \Vert^2 = \Vert x – y + y \Vert^2 \le (\Vert x – y \Vert + \Vert y \Vert)^2 = \Vert x – y \Vert^2 + 2 \Vert x – y \Vert \, \Vert y \Vert + \Vert y \Vert^2 \\
\iff \Vert x \Vert^2 – \Vert y \Vert^2 \le \Vert x – y \Vert^2 + 2 \Vert x – y \Vert \, \Vert y \Vert
$$
$$
\Vert y \Vert^2 = \Vert y – x + x \Vert^2 \le (\Vert y – x \Vert + \Vert x \Vert)^2 = \Vert x – y \Vert^2 + 2 \Vert x – y \Vert \, \Vert x \Vert + \Vert x \Vert^2 \\
\iff \Vert x \Vert^2 – \Vert y \Vert^2 \ge -\Vert x – y \Vert^2 – 2 \Vert x – y \Vert \, \Vert x \Vert
$$
At this point, I do not know how to convert the bounds including $\Vert x \Vert$, $\Vert y \Vert$ into anything useful.
Best Answer
In $\Bbb R$, consider $x=y+1$ with $y>0$. Then $\|x-y\|=1$, but $\bigl|\|x\|^2-\|y\|^2\bigr|=2y+1$ can be arbitrarily large.
However, we do have $$\|x\|^2-\|y\|^2=(\|x\|-\|y\|)(\|x\|+\|y\|)$$ if that helps you.