<s>[INST] Find all real solutions $x$ for the equation $x^{1/2} − (2−2x)^{1/2} = 1$ This is what the answer says:
Note that the equation can be rewritten as $\sqrt{x} − \sqrt{2 − 2x} = 1$,
and the existence of such real $x$ implies that $x$ is larger than or equal to $0$ and $x$ is less than or equal to $1$, since we implicitly assume that $\sqrt{x}$ and $\sqrt{2 − 2x}$ are real as well. Then
$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$, and
$\sqrt{x} − \sqrt{2 − 2x} = 1 \text{ if } x = 1$.
Thus, $x = 1$ is the only real solutions of the equation $\sqrt{x} − \sqrt{2 − 2x} = 1$.

I don't understand how
$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1  \text{ if }  x < 1$.
I understand everything else.
Can someone please explain? Thank you.
 [/INST] Notice that $\sqrt{x}-\sqrt{2-2x} \leq \sqrt{x}$. Then if $x<1$ we have $\sqrt{x}<\sqrt{1}=1$. It follows that $\sqrt{x}-\sqrt{2-2x} <1$ for $x<1$. So if $x<1$ it is not a solution. If $x=1$ we have $\sqrt{x}-\sqrt{2-2x}=\sqrt{1}-\sqrt{2-2}=1 $. So $x=1$ is a solution
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