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 $x \leq \frac{4}{3}$.

\begin{align*}
&\Rightarrow\sqrt{4-2x}\\
&=\sqrt{2+2(1-x)}&(x\in[0,2])\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{(1-x)}}\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\color{red}{\sqrt{(1-x)^2}}}&(\color{red}{x\in[0,1]})\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-(2x-x^2)}}\\
&=\sqrt{[1-\sqrt{2x-x^2}]+[1+\sqrt{2x-x^2}]+2\sqrt{1-\sqrt{2x-x^2}}\sqrt{1+\sqrt{2x-x^2}}}\\
&=\sqrt{\left(\sqrt{1-\sqrt{2x-x^2}}+\sqrt{1+\sqrt{2x-x^2}}\right)^2}\\
&=\sqrt{1-\sqrt{2x-x^2}}+\sqrt{1+\sqrt{2x-x^2}}
\end{align*}
