<s>[INST] Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.
Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.
 [/INST] The square roots are all well defined real numbers if and only if $0\le x\le2$. In this range, the numbers on both sides are positive, so we can say that
$$\begin{align}
LHS=RHS
&\iff\left(\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\right)^2=\left(\sqrt{4-2x}\right)^2\\
&\iff2+2\sqrt{1-(2x-x^2)}=4-2x\\
&\iff1+\sqrt{(1-x)^2}=2-x\\
&\iff|1-x|=1-x\quad\text{(note the absolute values!)}\\
&\iff1-x\ge0\\
&\iff x\le1
\end{align}$$
Together with the restriction $0\le x\le2$, we have
$$\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}=\sqrt{4-2x}\iff 0\le x\le1$$
Remark: If it seems odd that the identity suddenly stops working at $x=1$, note that what we really have is the two-piece identity
$$\sqrt{4-2x}=
\begin{cases}
\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }0\le x\le1\\
\sqrt{1+\sqrt{2x-x^2}}-\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }1\le x\le2
\end{cases}$$
</s>