<s>[INST] Square root inequality $\sqrt {x-z} \geq \sqrt x -\sqrt{z} $ Does the following inequality hold?
$$\sqrt {x-z} \geq \sqrt x -\sqrt{z} \  , $$
for all $x \geq z \geq 0$.
My justification
\begin{equation}
z \leq x \Rightarrow \\ \sqrt z \leq \sqrt {x} \Rightarrow \\ 2\sqrt z \sqrt z \leq 2\sqrt z\sqrt {x} \Rightarrow \\ 2 z \leq 2\sqrt z\sqrt {x} \Rightarrow  \\ z - 2\sqrt z\sqrt {x} + x \leq x - z  \Rightarrow  \\  (\sqrt x -\sqrt z )^2 \leq x - z \Rightarrow  \\ \sqrt x -\sqrt z \leq \sqrt {x - z}
\end{equation}
 [/INST] \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}
</s>