<s>[INST] Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
 [/INST] Well assume that $\sqrt{a+\sqrt{b}}$ can be written as sum of 2 square roots
$$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}\\a+\sqrt{b}=x+y+\sqrt{4xy}\\a=x+y\\b=4xy\\x=a-y\\b=4(a-y)y\\b=4ay-4y^2\\4y^2-4ay+b=0\\y_{1,2}=\frac{4a\pm\sqrt{16a^2-16b}}{8}\\y_{1,2}=\frac{a\pm\sqrt{a^2-b}}{2}\\x_{1,2}=\frac{a\mp\sqrt{a^2-b}}{2}$$
Now it $x_1=y_2$ and $x_2=y_1$ so that doesn't matter at all,now set $C=\sqrt{a^2-b}$ and you get your formula
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