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.
 A: 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.
A: I suggest rewriting this in a form that looks more general, first by noticing that $z - x$ gets repeated, and then by noticing that the denominator on the first squared expression of the RHS is the sum of the denominators of the other two squared expression (the squared expression of the LHS has denominator 1). Then the inequality becomes
$$\frac{|u|^2}{a} + \frac{|v|^2}{b} \ge \frac{|u + v|^2}{a + b}$$
by replacing $u$ for $z - x - y$, $v$ for $y$, $a$ for 1, and $b$ for $c - 1$.
The LHS is $\ge \frac{(|u| + |v|)^2}{a + b}$ due to the Cauchy-Schwarz inequality (look for Sedrakyan's lemma), since $a, b > 0$. $\frac{(|u| + |v|)^2}{a + b} \ge$ the RHS due to the triangle inequality.
Edit 1: Fixed mixed up LHS and RHS in last paragraph. Removed superfluous observation about $|u|,|v|≥0$.
Edit 2: Following Brian Chao's solution, we may also compute the difference between subtract the RHS from the LHS to obtain
$$|u|^2\frac{b}{a(a+b)} + |v|^2\frac{a}{b(a + b)} - 2\langle u,v\rangle\frac{1}{a+b} = \Big|u\sqrt{\frac{a}{b(a + b)}} + v\sqrt{\frac{b}{a(a + b)}}\Big|^2$$
