<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] You may first symmetrise $x(2-x)$ by setting $\boxed{x=y+1}$:
$$\sqrt{1+\sqrt{1-y^2}} + \sqrt{1-\sqrt{1-y^2}}= \sqrt{2(1-y)} \text{ with } -1\leq y\leq 1$$
Now, squaring gives
$$2+2|y|= 2(1-y) \Leftrightarrow |y| = -y$$
Hence, with $-1\leq y\leq 1 \stackrel{|y|=-y}{\Rightarrow} -1\leq y\leq 0 \stackrel{x=y+1}{\Rightarrow} \boxed{0 \leq x \leq 1}$ is the solution.
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