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?
 A: The number $\sqrt{1-x}-\sqrt{x}$ can be negative and, in this case, the inequality doesn't hold. Squaring both sides can be done only if both sides are nonnegative. Thus your inequality becomes
$$
\begin{cases}
0\le x\le 1 \\[8px]
\sqrt{1-x}-\sqrt{x}\ge0 \\[4px]
1-x-2\sqrt{x(1-x)}+x>\dfrac{1}{3}
\end{cases}
$$
The second inequality becomes $1-x\ge x$, so $x\le1/2$. The third inequality is
$$
\sqrt{x(1-x)}<\frac{1}{3}
$$
so $9x^2-9x+1>0$ that has the solutions
$$
x<\frac{3-\sqrt{5}}{6}\qquad\text{or}\qquad x>\frac{3+\sqrt{5}}{6}
$$
that should be combined with $0\le x\le 1/2$. Thus the solution set of the original equation is
$$
\left[0,\frac{3-\sqrt{5}}{6}\right)
$$
because $(3+\sqrt{5})/6>1/2$.

A different strategy is to write the inequality as
$$
\sqrt{3}\sqrt{1-x}>1+\sqrt{3}\sqrt{x}
$$
With the constraint $0\le x\le1$, this becomes
$$
3-3x>1+2\sqrt{3}\sqrt{x}+3x
$$
or
$$
3x+\sqrt{3}\sqrt{x}-1<0
$$
Setting $t=\sqrt{x}$ (with $0\le t\le1$), this is quadratic: $3t^2+\sqrt{3}t-1<0$. So we have the solution set
\begin{cases}
0\le t \le 1 \\[8px]
\dfrac{-\sqrt{3}-\sqrt{15}}{6}<t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}
\end{cases}
that becomes
$$
0\le t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}
$$
and therefore
$$
0\le x<\left(\dfrac{-\sqrt{3}+\sqrt{15}}{6}\right)^2=\frac{3-\sqrt{5}}{6}
$$
A: Observe first that if must be $\;0\le x\le 1\;$ , so the given interval is within this constraint. Second, square:
$$\sqrt{1-x}-\sqrt x\ge\frac1{\sqrt3}\implies1-x-2\sqrt{x(1-x)}+x\ge\frac13\implies$$
$$2\sqrt{x(1-x)}\le\frac23\implies x(1-x)\le\frac19\implies 9x^2-9x+1\ge0\implies$$
The last quadratic's roots are given by
$$x_{1,2}=\frac{9\pm\sqrt{45}}{18}=\frac{3\pm\sqrt5}6\implies$$
$$ 9x^2-9x+1\ge0\iff \left(x-\frac{3-\sqrt5}6\right)\left(x-\frac{3+\sqrt5}6\right)\ge0\iff$$
$$x\le\frac{3-\sqrt5}6\;\;\text{or}\;\;x\ge\frac{3+\sqrt5}6\;\;\;\color{red}{(***)}$$
and  we get what we want from the leftmost inequality.
Added on request: If we take
$$f(x)=\sqrt{1-x}-\sqrt x\implies f'(x)=-\frac1{2\sqrt{1-x}}-\frac1{2\sqrt x}=-\frac12\left(\frac1{\sqrt{1-x}}+\frac1{\sqrt x}\right)<0\implies$$
$\;f\;$ is monotone descending, and since 
$$f\left(\frac{3-\sqrt5}6\right)=\sqrt{1-\frac{3-\sqrt5}6}-\sqrt\frac{3-\sqrt5}6=\frac1{\sqrt6}\left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)\le\frac1{\sqrt3}\;(*)$$
since
$$(*)\iff \left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)^2\le\left(\sqrt\frac63\right)^2\iff6-4\le2\;\;\color{green}\checkmark$$
so in $\;\color{red}{(***)}\;$ only the left inequality applies.
A: Here is a solution mostly based on a graphical understanding (nevertheless written with rigorous arguments). 
We acknowledge first that the domain of definition of the inequation is $(0,1)$.
Let us write this inequation under the form:
$$\sqrt{1-x}>\dfrac{1}{\sqrt{3}}+\sqrt{x}$$
As it is an inequality between positive numbers, it is equivalent to the "squared" inequality:
$$1-x>\dfrac{1}{3}+\dfrac{2}{\sqrt{3}}\sqrt{x}+x$$
Re-arranging this inequality in the following way:
$$\tag{1}\dfrac{1}{3}-x>\dfrac{1}{\sqrt{3}}\sqrt{x}$$
(1) can be expressed under the form: $\tag{2}f(x)>g(x)$ where


*

*$f(x):=\dfrac{1}{3}-x$, decreasing on $[0,1]$ with, in particular, $f(0)=1/3$ and $f(1/3)=0.$

*$g(x):=\dfrac{1}{\sqrt{3}}\sqrt{x}$, increasing on $[0,1]$ with, in particular, $g(0)=0$ to $g(1/3)=\dfrac{1}{3}>0.$
Thus, their associated curves possess a unique intersection point $A$ whose abscissa $x_0$ is the solution to $f(x)=g(x)$ (see figure below). A little computation (already done by the OP) shows that $x_0$ is one of the roots $x_0=\dfrac{3\pm\sqrt{5}}{6}$ of 
$$\tag{3} 9x^2-9x+1=0$$
This root has to be $<1/3$ ; thus it is $x_0=\dfrac{3-\sqrt{5}}{6}$.
Thus the looked for region corresponds to the values of $x$ that are in $[0,x_0)$ (See curves below). 
Remark: an interesting fact is that the "fake" second solution of the quadratic equation (3) can be visualized (and then understood) as the intersection $B$ of the straight line with "mirror" (dotted) curve with equation $y=-\sqrt{x/3}$, which appears into the play when $\sqrt{x}$ is squared... 

A: 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
$[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)] \cap [0,1/2)$.
=====
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 $x \in (\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)\cap [0,1/2]= [0,\frac{3-\sqrt{5}}{6})$
