Example 6. Given $x_{1}, x_{2}, x_{3}>0$, and $x_{1}+x_{2}+x_{3}=1$. Find
$$
u=\sqrt{1-x_{1}}+\sqrt{1-x_{2}}+\sqrt{1-x_{3}}
$$

the upper and lower bounds.
Consider three points in the Cartesian coordinate system:
$$
\begin{array}{l}
A\left(x_{1}, \sqrt{1-x_{1}}\right), B\left(x_{2}, \sqrt{1-x_{2}}\right), \\
C\left(x_{3}, \sqrt{1-x_{3}}\right).
\end{array}
$$

It is evident that they all lie on the parabola $y=\sqrt{1-x}$. As shown in Figure 9, let the centroid of $\triangle A B C$ be $\boldsymbol{G}\left(x_{0}, y_{0}\right)$, then
$$
\begin{aligned}
x_{0} & =\frac{x_{1}+x_{2}+x_{3}}{3}=\frac{1}{3} \\
y_{0} & =\frac{\sqrt{1-x_{1}}+\sqrt{1-x_{2}}+\sqrt{1-x_{3}}}{3} \\
& =\frac{u}{3}.
\end{aligned}
$$

Clearly, $G$ is below the parabola and above the chord $E F$, which means
$$
H Q < y_{0} = H G S H P.
$$

It is easy to calculate
$$
H Q=\frac{2}{3}, H P=\sqrt{1-\frac{1}{3}}=\frac{\sqrt{6}}{3},
$$

Thus, $\frac{2}{3} \leqslant y=\frac{u}{3} \leqslant \frac{\sqrt{6}}{3}$,
which implies that $\sqrt{6}$ and $2$ are the upper and lower bounds of $u$, respectively. Therefore, the original inequality (7) can be improved to
$$
2 < \sqrt{1-a^{2}} + \sqrt{1-b^{2}} + \sqrt{1-c^{2}}
$$
$< \sqrt{6}$.
This once again demonstrates the ingenuity of the geometric proof, which lies in the skillful selection of appropriate figures.
(Author's affiliation: Department of Mathematics, Guizhou Education College)