<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] Note
$$LHS^2= \left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} \right)^2\\
=2+2\sqrt{(1 -x)^2}= {4-2x}=RHS^2
$$
So, the solution is just the domain of the equation, which is jointly determined by
$$2x-x^2\ge 0,\>\>\>1-\sqrt{2x-x^2}\ge 0,\>\>\>4-2x\ge 0$$
which leads to $0\le x\le 1$.
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