<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 domain gives $0\leq x\leq2$.
Now,  $$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{\left(\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}}\right)^2}=$$
$$=\sqrt{2+2\sqrt{1-2x+x^2}}=\sqrt{2+2|x-1|}.$$
Thus, it's enough to solve $$1+|x-1|=2-x.$$
Can you end it now?
I got $0\leq x\leq1$.
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