Quick inequality verification The assertion is:

Let $c>1$. Then it holds that
$$\forall x,y,z \in \mathbb{R}^n,\;
\left|z-x-y\right|^2 \ge \dfrac{\left|z-x\right|^2}{c}
 - \dfrac{\left|y\right|^2}{c-1}$$

I can not think of anything more than maybe using the reverse triangle inequality, although it is not leading me anywhere until now. Thanks in advance.

Let us prove the inequality without using the triangle inequality. For $x,y\in\mathbb{R}^n$, we write $\langle x,y\rangle=\sum_{i=1}^{n}x_iy_i$ for the standard inner product of $x$ and $y$. First observe that by substituting $z-x$ with $x$, it suffices to prove
$$\forall x,y\in\mathbb{R}^n,\enspace|x-y|^2\geq \frac{|x|^2}{c}-\frac{|y|^2}{c-1}.$$
Equivalently,
$$\forall x,y\in\mathbb{R}^n,\enspace|x-y|^2-\frac{|x|^2}{c}+\frac{|y|^2}{c-1}\geq 0.$$
The left-hand side equals
$$\begin{align*}|x|^2+|y|^2-2\langle x,y\rangle-\frac{|x|^2}{c}+\frac{|y|^2}{c-1}&=\frac{c-1}{c}|x|^2+\frac{c}{c-1}|y|^2-2\langle x,y\rangle\\&=\Big|\sqrt{\frac{c-1}{c}}x-\sqrt{\frac{c}{c-1}}y\Big|^2,\end{align*}$$
which is always nonnegative. This proves the inequality.
