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.

Observe that
$$\sqrt{x} - \sqrt{2-2x} = 1 \implies x = 1 + 2-2x + 2\sqrt{2-2x} \implies 3x - 3 = \sqrt{2-2x}$$ $$ \implies 9(x^2 - 2x + 1) = 2-2x \implies 9x^2 - 16x +7 = 0 \implies (x -1)(9x - 7) = 0  $$
Thus $ x = 1, \frac{7}{9} $. Now, $$ \sqrt{\frac{7}{9}}-\sqrt{2-\frac{14}{9}} \neq 1 $$
Thus $x=1 $ is the solution.
