<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] 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: $$2u = \sqrt{2(A+C)} + \sqrt{2(A-C)}$$
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