Square Root of Sum as Sum of Square Roots/Proof 2

Theorem

Let $a, b \in \R, a \ge b$.

Then:

$\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$

$\sqrt p + \sqrt q = \sqrt {p + q + \sqrt {4pq}}$

Let

$p = \dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2$,

$q = \dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2$.

Then

$\ds p + q$

$=$

$\ds \frac a 2 + \frac {\sqrt {a^2 - b^2} } 2 + \frac a 2 - \frac {\sqrt {a^2 - b^2} } 2$

$\ds $

$=$

$\ds a$

$\ds \sqrt {4pq}$

$=$

$\ds \sqrt {4 \paren {\frac a 2 + \frac {\sqrt {a^2 - b^2} } 2} \paren {\frac a 2 - \frac {\sqrt {a^2 - b^2} } 2} }$

$\ds $

$=$

$\ds \sqrt {\paren {a + \sqrt {a^2 - b^2} } \paren {a - \sqrt {a^2 - b^2} } }$

$\ds $

$=$

$\ds \sqrt {a^2 - \paren {a^2 - b^2} }$

$\ds $

$=$

$\ds \sqrt {b^2}$

$\ds $

$=$

$\ds b$

Therefore:

$\sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2} = \sqrt {a + b}$

$\blacksquare$