Proving $\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$, where $a\geq 0,\ b\geq 0$. Who can help me to prove this equation?
$$\sqrt{2a + 2\sqrt{a^2 - b}}=\sqrt{a-\sqrt{b}} + \sqrt{a+\sqrt{b}}$$ Where, $$a\geq 0,\ b\geq 0$$

Notice, here is the correct proof $$LHS=\sqrt{2a+2\sqrt{a^2-b}}$$
$$=\sqrt{(a-\sqrt{b})+(a+\sqrt b)+2\sqrt{a^2-(\sqrt b)^2}}$$
$$=\sqrt{\underbrace{\left(\sqrt{a-\sqrt{b}}\right)^2}_{A^2}+\underbrace{\left(\sqrt{a+\sqrt b}\right)^2}_{B^2}+2\underbrace{\sqrt{(a-\sqrt b)}}_{A}\underbrace{\sqrt{(a+\sqrt b)}}_{B}}$$
using identity, $A^2+B^2+2AB=(A+B)^2$, 
$$=\sqrt{\left(\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}\right)^2}$$
$$=\left|\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}\right|$$
since, $a\ge 0,\ b\ge 0 $
$$=\sqrt{a-\sqrt b}+\sqrt{a+\sqrt b}$$
$$=RHS$$
