Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?

Alternative approach:
If $a,b > 0$ and $a^2 = b^2$, then $a$ must $= b$.
Let $a = \frac{1}{2} \left(\sqrt{6} + \sqrt{14}\right),
~b = \sqrt{5 + \sqrt{21}} \implies 0 < a,b.$
Then $a^2 = \frac{1}{4} \left(6 + 14 + 2\sqrt{84}\right)
= 5 + \sqrt{21} = b^2.$
