<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] Before starting, let us note that the formula is not really $a$ nested
radicals formula, because it suffices to replace $\sqrt{B}$ by $B,$ and
replace $C^{2}=A^{2}-B,$ by $C^{2}=A^{2}-B^{2}.$ So let me show you how starting
from the l.h.s. written as
\begin{equation*}
\sqrt{A+B}
\end{equation*}
to arrive to the r.h.s written as
\begin{equation*}
\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ \ with\ \ \ C=\sqrt{%
A^{2}-B^{2}}.
\end{equation*}
Consider a rectangle triangle with sides $A$, $B$, and $C$ such that
(pythagor theorem) 
\begin{equation*}
A^{2}=B^{2}+C^{2}\ \ \ \ \ \ (\ast )
\end{equation*}
then
\begin{eqnarray*}
B^{2} &=&A^{2}-C^{2} \\
B^{2} &=&(A+C)(A-C) \\
B^{2} &=&4\frac{(A+C)}{2}\frac{(A-C)}{2}
\end{eqnarray*}
then
\begin{equation*}
B=2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}
\end{equation*}
and by adding $A$ to both sides
\begin{equation*}
A+B=A+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}
\end{equation*}
However, note that
\begin{equation*}
A=\frac{A+C}{2}+\frac{A-C}{2}
\end{equation*}
then
\begin{eqnarray*}
A+B &=&A+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\
&=&\frac{A+C}{2}+\frac{A-C}{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\
&=&\left( \sqrt{\frac{A+C}{2}}\right) ^{2}+\left( \sqrt{\frac{A-C}{2}}%
\right) ^{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}} \\
A+B &=&\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right) ^{2}
\end{eqnarray*}
and then
\begin{equation*}
\sqrt{A+B}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
\end{equation*}
where $C$ is given by (*), that is, $C=\sqrt{A^{2}-B^{2}}.$
</s>