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

$\sqrt{a}$ 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 
$$|\sqrt{a}-\sqrt{b}|^2\leq |a-b|$$
