Square Root of Number Plus Square Root

Theorem
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.

Then:
 * $\ds \sqrt {a + \sqrt b} = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}$