<s>[INST] 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?
 [/INST] You're probably missing a $+$

The key is to convert it to the form $a^2+2ab+b^2$. Here you have $\sqrt{21}$ inside the outer square root, so first multiply and divide by $\sqrt 2$. Now $a$ and $b$ can be found easily as $7+3 = 10$ and $7\cdot 3 = 21$

$\begin{align}\sqrt{5+\sqrt{21}}& = \frac1{\sqrt2}\sqrt{10+2\sqrt{21}} = \frac1{\sqrt2}\sqrt{(7+3)+2\sqrt{7\cdot3}}\\& = \frac1{\sqrt2}\sqrt{(\sqrt7+\sqrt3)^2} = \frac1{\sqrt2}(\sqrt 7+\sqrt 3)\\& =\frac1{\sqrt2\cdot\sqrt2}(\sqrt {14}+\sqrt 6)\\&=\frac1{2}(\sqrt 6+\sqrt {14})\end{align}$
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