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.
 A: 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
A: 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, $C=\sqrt{A^{2}-B^{2}}.$
A: In what follows, I will show how to start from the l.h.s. written as
\begin{equation*}
\sqrt{A+\sqrt{B}}
\end{equation*}
and 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}.
\end{equation*}
To this end, make use of the following standard identities of algebra 
\begin{eqnarray*}
x^{2}-y^{2} &=&(x+y)(x-y) \\
(x+y)^{2} &=&x^{2}+y^{2}+2xy
\end{eqnarray*} 
Note that this second identity can be written as follows 
\begin{equation}
xy=\frac{1}{2}(x+y)^{2}-\frac{1}{2}x^{2}-\frac{1}{2}y^{2}  \tag{K}
\end{equation} 
Consider a rectangle triangle with the longest side $A$ and the others are $ 
\sqrt{B,}$ and $C.$ So, by the Pythagorean theorem one has,$\ $ 
\begin{equation}
A^{2}=B+C^{2},  \tag{P}
\end{equation}
then$\ \sqrt{B}=\sqrt{A^{2}-C^{2}},$ therefore, 
\begin{eqnarray*}
\sqrt{A+\sqrt{B}} &=&\sqrt{A+\sqrt{A^{2}-C^{2}}} \\
&=&\sqrt{A+\sqrt{(A+C)(A-C)}} \\
&=&\sqrt{A+\sqrt{A+C}\sqrt{A-C}},\ let\ x=\sqrt{A+C},\ and\ y=\sqrt{A-C},\
and\ apply\ (K) \\
&=&\sqrt{A+\frac{1}{2}\left( \sqrt{A+C}+\sqrt{A-C}\right) ^{2}-\frac{1}{2}%
\left( \sqrt{A+C}\right) ^{2}-\frac{1}{2}\left( \sqrt{A-C}\right) ^{2}} \\
&=&\sqrt{A+\frac{1}{2}\left( \sqrt{2}\sqrt{\frac{A+C}{2}}+\sqrt{2}\sqrt{%
\frac{A-C}{2}}\right) ^{2}-\frac{1}{2}(A+C)-\frac{1}{2}\left( A-C\right) } \\
&=&\sqrt{A+\frac{1}{2}(\sqrt{2})^{2}\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{%
A-C}{2}}\right) ^{2}-A} \\
&=&\sqrt{\left( \sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right) ^{2}} \\
\sqrt{A+\sqrt{B}} &=&\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}},\ \ \ with\ C=%
\sqrt{A^{2}-B},\ from\ (P).
\end{eqnarray*}
A: $$A+\sqrt{B}=\frac{A+C}{2}+\frac{A-C}{2}+\sqrt{A^2-(A^2-B)}$$
$$={\sqrt{\frac{A+C}{2}}}^2+2\sqrt{\frac{(A+\sqrt{A^2-B})(A-\sqrt{A^2-B})}{4}}+{\sqrt{\frac{A-C}{2}}}^2$$
$$={\sqrt{\frac{A+C}{2}}}^2+2\sqrt{\frac{(A+C)(A-C)}{4}}+{\sqrt{\frac{A-C}{2}}}^2$$
$$=\left(\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right)^2$$
A: $$A+\sqrt{B}=\frac{A}{2}+\frac{C}{2}+\frac{A}{2}-\frac{C}{2}+\sqrt{B},$$
where $C=\sqrt{A^2-B}$. But
$$\sqrt{B}=\sqrt{A^2-(A^2-B)}=\frac{2}{2}\sqrt{(A+\sqrt{A^2-B})(A-\sqrt{A^2-B})}=2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}.$$
Then
$$A+\sqrt{B}=\frac{A+C}{2}+\frac{A-C}{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}=\left(\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right)^2.$$
Taking square roots gives the answer.
A: Here's an awesome proof I discovered right now :
Consider the quadratic :
$x^2-Ax+\frac{B}{4}=0$ with the roots $x_1$ and $x_2$ .
I have chosen this because of the nice determinant $A^2-4 \cdot \frac{B}{4}=A^2-B=C^2$
Now let's look at the roots :
$$x_{1,2}=\frac{A \pm C}{2}$$ wow exactly those numbers.
Now use Viete's relations to get :
$x_1+x_2=A$ and $x_1\cdot x_2=\frac{B}{4}$ 
The problem is now equivalent with :
$$\sqrt{x_1}+\sqrt{x_2}=\sqrt{x_1+x_2+2\sqrt{x_1x_2}}$$ which should be obvious .
This is very simple but I felt a great joy when I found it . This is the beauty of mathematics :D.
A: I can derive the left side from the right, not the other way around.
The right side is the sum of the two positive roots (assuming $A^2>B$) of:
$$x^4-Ax^2+\frac{B}{4}=0$$
But this polynomial factors as:
$$x^4-Ax^2+\frac{B}{4}=\left(x^2+\frac{\sqrt{B}}2\right)^2-(A+\sqrt{B})x^2\\
=
\left(x^2-\sqrt{A+\sqrt{B}}x+\frac{\sqrt{B}}2\right)
\left(x^2+\sqrt{A+\sqrt{B}}x+\frac{\sqrt{B}}2\right)$$
Note that the positive values cannot be roots to the right factor, so they have to be roots of the left factor, and therefore their sum is $\sqrt{A+\sqrt{B}}$.

Reversing the direction is uglier:
$$A+\sqrt{B}$$ is the larger root of:
$$x^2-2Ax + C^2 = 0$$
Therefore $w=\sqrt{A+\sqrt{B}}$ is the largest root of $$x^4-2Ax^2+C^2 = 0$$
Now $$x^4-2Ax^2+ C^2= \left(x^2+C\right)^2-2(A+C)x^2\\
=\left(x^2-\sqrt{2(A+C)}x+C\right)\left(x^2+\sqrt{2(A+C)}x+C\right)$$
It can't be a root of the right side, which is positive at $w$, so we have:
$$w^2-\sqrt{2(A+C)}w+C=0$$
We can also factor the above by completing the square the other way:
$$x^4-2Ax^2+C^2 = (x^2-C)^2 - 2(A-C)x^2 =\\
=\left(x^2-\sqrt{2(A-C)}x-C\right)\left(x^2+\sqrt{2(A-C)}x-C\right)$$
The positive roots of this polynomial have to be split on the two sides, and we can see since that $\sqrt{A+\sqrt{B}}$ must be the largest root, it must be a root of the left factor.
A common root $u=\sqrt{A+\sqrt{B}}$ of the two equations:
$$w^2 - \sqrt{2(A+C)}w + C=0\\
w^2-\sqrt{2(A-C)}w - C=0$$
Thus, adding, then dividing by $w$ and re-arranging, we get: $$2u = \sqrt{2(A+C)} + \sqrt{2(A-C)}$$
