Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate  $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my lights go off.
Any leads? Thanks in advance.

Although it is not clear what range of values is acceptable for the variables, I believe I have a sense of what you are trying to do here. This is called denesting square roots.
In particular, if $a,b,c$ are positive real numbers such that $a^2-b^2 c$ is non-negative, then it holds that
$$\sqrt{a+ b\sqrt{c}}=\sqrt{\frac{a+\sqrt{a^2-b^2 c}}{2}}+ \sqrt{\frac{a-\sqrt{a^2-b^2 c}}{2}}.$$
It can be easily proven by squaring both sides and using the difference of squares factorization. There is a way of coming up with it naturally by using the same idea behind finding complex square roots, though it might not be an entirely rigorous approach.
This yields
\begin{align*}
\sqrt{a+b+\sqrt{2ab+b^2}}&=\sqrt{\frac{a+b+|a|}{2}}+\sqrt{\frac{a+b-|a|}{2}}\\
&= \begin{cases} \sqrt{a+\frac{b}{2}}+\sqrt{\frac{b}{2}} &\text{ if } a\ge 0\\ \sqrt{\frac{b}{2}}+\sqrt{a+\frac{b}{2}} &\text{ if } a<0\end{cases}.
\end{align*}
which works because $$(a+b)^2-1^2\cdot (2ab+b^2)=a^2$$ is non-negative. Both cases yield the same answer.
