Example 8. Solve the inequality $\sqrt{1-x}-\sqrt{x}>\frac{1}{\sqrt{3}}$.
Solve for the set of permissible values of the unknown $\mathrm{x}$ as $0 \leqslant \mathrm{x} \leqslant 1$.
Let $\mathrm{y}_{1}=\sqrt{1-\mathrm{x}}, \mathrm{y}_{2}=\sqrt{\mathrm{x}}+\frac{1}{\sqrt{3}}$, then the graphs of $\mathrm{y}_{1}, \mathrm{y}_{2}$ are shown in Figure 2.

Now find the x-coordinate of the intersection point $\mathrm{A}$ of the parabolas $\mathrm{y}_{1}$ and $\mathrm{y}_{2}$. Solve the equation $\mathrm{y}_{1}=y_{2}$, that is
$\sqrt{1-\mathrm{x}}=\sqrt{\mathrm{x}}+\frac{1}{\sqrt{3}}$,

Square both sides and rearrange, we get

Square both sides again, and rearrange to get
$$
x^{2}-x+\frac{1}{9}=0 .
$$

Solving it, we get
$$
\mathrm{x}_{1}=\frac{3-\sqrt{5}}{6}, \quad \mathrm{x}_{2}=\frac{3+\sqrt{5}}{6}
$$

Substituting $x_{2}$ into equation (2), the left side is negative while the right side is positive, so it is an extraneous root. Substituting $\mathrm{x}_{1}$ directly into equation (1), both sides are equal, so the x-coordinate of the intersection point $\mathrm{A}$ of the two curves is $\mathrm{x}=\frac{3-\sqrt{5}}{6}$. Combining with Figure 2, it can be seen that when $0 \leqslant \mathrm{x}<\frac{3-\sqrt{5}}{6}$, the solution to $y_{1} < y_{2}$ is $0 \leqslant \mathrm{x}<\frac{3-\sqrt{5}}{6}$, which is the solution to the original inequality.