Prove the difference of roots is less than or equal to the root of the difference I am doing a larger proof that requires this to be true:
$|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|}$
I can square both sides to get
$a - 2\sqrt{a}\sqrt{b} + b \leq |a - b|$
Note that a and b are > 0.
I also know that
$|c| - |d| \leq |c - d|$
It seems like a mistake that $+b$ is there on the left...
How can I prove this is true

We may suppose that $a\ge b$. So,
$$\begin{align}a-2\sqrt{ab}+b\le|a-b|&\iff a-2\sqrt{ab}+b\le a-b\\&\iff b\le \sqrt {ab}\\&\iff b^2\le ab\\&\iff b(a-b)\ge 0.\end{align}$$
Hence, all we need is to prove that $b(a-b)\ge 0$ holds for any $a\ge b\ge 0$ and it does hold for any $a\ge b\ge 0$. So, $a-2\sqrt{ab}+b\le|a-b|$ also holds for any $a\ge b\ge 0$.
