<s>[INST] Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which yields:
$$\frac{1}{3} > \sqrt{x}\sqrt{1 -x}.$$ 
Squaring it again we get inequality:
$$9x^2 - 9x + 1 > 0,$$
solution to which is a domain $[0, \frac{3 - \sqrt{5}}{6}) \cup (\frac{3 + \sqrt{5}}{6}, 1]$. But, solution to original inequality is just $[0, \frac{3 - \sqrt{5}}{6})$. What am I omitting/where I'm doing mistakes?
 [/INST] Well, there is this but I'm not sure how to generalize it.
$\sqrt{1-x} - \sqrt{x} > 0$ so $1-x > x$ so $x < \frac 12$ and $x \in [0,1/2)$.
When you square but sides of $\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$ to get $\frac{1}{3} > \sqrt{x}\sqrt{1 -x}$ you are extraneously adding the possibility that $\sqrt{x} - \sqrt{1-x} > 1/\sqrt{3}$ (i.e that maybe $x \in (1/2, 1]$) which we know can not be true.
When we square a second time to get $9x^2 - 9x + 1 > 0$ we are extraneously adding the possiblities that $x < 0$ or $x > 1$.  (You caught those).
The solution to  $9x^2 - 9x + 1 > 0$ is $(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)$.
You caught it should be $[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)]\cap [0,1]$ but you didn't catch it should be
$[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)] \cap [0,1/2)$.
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I guess the way to generalize this to note every time you square an inequality make a note to combine with the known inequality of sign.
i.e.
$\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$
Square both sides and conclude
$\sqrt{x}\sqrt{1-x} < 1/3$  AND $\sqrt{1-x} - \sqrt{x} \ge 0$.
So $\sqrt{1-x} \ge \sqrt{x}$.
Square those both side to get $1-x \ge x$ AND $x \ge 0$ and $x \le 1$
So $\sqrt{x}\sqrt{1-x} < 1/3$ And $x \in (-\infty, 1/2]) \cap [0,\infty) \cap (-\infty, 1] = [0,1/2]$.
Square both sides to get
$9x^2 - 9x + 1 > 0$ AND $\sqrt{x}\sqrt{1-x} \ge 0$ and $x \in [0,1/2]$
So $9x^2 - 9x + 1 > 0$ and $x \in [0,\infty) \cap (-\infty,1] \cap [0,1/2] = [0,1/2]$. 
So $x \in (\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)\cap [0,1/2]= [0,\frac{3-\sqrt{5}}{6})$
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