Square root inequality $\sqrt {x-z} \geq \sqrt x -\sqrt{z} $

More simply we have that

$$x-z=\left(\sqrt x -\sqrt{z}\right)\left(\sqrt x +\sqrt{z}\right)\ge \left(\sqrt x -\sqrt{z}\right)^2$$

While the other existing answers give simple algebraic reasons for this fact, it is actually far more useful in general to see this fact as a special case of the smoothing technique. In particular, for any concave function $f$ on domain $D⊆ℝ$, we have that $f(a+b) ≥ f(a'+b')$ for every $a,b,a',b'$ such that $a+b = a'+b'$ and $a' ≤ a,b ≤ b'$. That is, pushing the points $a,b$ apart while preserving their sum decreases the total value of $f$ on them. In your case you simply have $(a,b) = (z,x-z)$ and $(a',b') = (0,x)$ and $f$ being the real-square-root function.

This general smoothing technique is extremely powerful if you know how to use it. For example it gives a one-line proof of AM-GM inequality, and similarly a short proof of Jensen's inequality. In discrete mathematics it is sometimes called a swapping argument (here is an example usage). In real analysis, it can be used in conjunction with a compactness argument to prove theorems that can be quite difficult to prove without (such as the two continuous optimization theorems in this post).

\begin{equation} \qquad\sqrt {x-z} \ge \sqrt x -\sqrt{z}\\ \implies (\sqrt {x-z})^2 \geq (\sqrt x -\sqrt{z})^2\\ \implies x-z\ge x-2\sqrt{xz}+z\\ \implies x-z - x-z\ge-2\sqrt{xz}\\ \implies -2z\ge -2\sqrt{xz}\\ \implies -z\ge -\sqrt{xz}\\ \text{ subtracting both sides from both sides reverses the relationship}\\ \implies \sqrt{xz}\ge z\\ \implies xz\ge z^2\\ \implies x\ge z\quad \land\quad x-z\ge 0\\ \implies x\ge\ z\ge 0 \end{equation}