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

Solution 1:

$\sqrt{a}$ and $\sqrt{a}$ are positive, therefore, $$-\sqrt{b}\leq \sqrt{b}$$ ant it follows that $$\sqrt{a}-\sqrt{b}\leq \sqrt{a} +\sqrt{b} $$

by symmetry we have $$|\sqrt{a}-\sqrt{b}|\leq |\sqrt{a} +\sqrt{b} |=\sqrt{a} +\sqrt{b}$$ Now muptiply both sides by $|\sqrt{a}-\sqrt{b}|$ to get $$|\sqrt{a}-\sqrt{b}|^2\leq |a-b|$$

Solution 2:

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 $a\ge b\ge 0$.