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 $x \leq \frac{4}{3}$.
 A: 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 $0\leq x\leq1$.
A: 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 $0\le x\le 1$.
A: \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*}
A: 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.
A: 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
$$\sqrt{4-2x}=
\begin{cases}
\sqrt{1+\sqrt{2x-x^2}}+\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }0\le x\le1\\
\sqrt{1+\sqrt{2x-x^2}}-\sqrt{1-\sqrt{2x-x^2}}\quad\text{for }1\le x\le2
\end{cases}$$
