［Example 5.4.2］Solve the inequality:
$$
\sqrt{x^{2}+6 x+10}+\sqrt{x^{2}-6 x+10}>10 .
$$
$$
\left\{\begin{array}{l}
x^{2}+6 x+10 \geqslant 0, \\
x^{2}-6 x+10 \geqslant 0, \\
\sqrt{x^{2}+6 x+10}>10-\sqrt{x^{2}-6 x+10} .
\end{array}\right.
$$

Inequalities (1) and (2) have solutions for all real numbers, so we only need to solve inequality (3).
(1) When $10-\sqrt{x^{2}-6 x+10} \leqslant 0$, inequality (3) is always true.

For this, we only need to solve
$$
\begin{array}{c}
10-\sqrt{x^{2}-6 x+10} \leqslant 0, \\
x^{2}-6 x-90 \geqslant 0, \\
(x-3)^{2} \geqslant 99, \\
x \geqslant 3+\sqrt{99} \text { or } x \leqslant 3-\sqrt{99} .
\end{array}
$$
(2) When $10-\sqrt{x^{2}-6 x+10}>0$, squaring both sides of inequality (3) gives
$$
\begin{array}{c}
x^{2}+6 x+10>100+x^{2}-6 x+10-20 \sqrt{x^{2}-6 x+10}, \\
\left\{\begin{array}{l}
25-3 x>0, \\
25\left(x^{2}-6 x+10\right)>(25-3 x)^{2}
\end{array} \text { or } 25-3 x \leqslant 0 .\right. \\
x<\frac{5}{4} \sqrt{15} \text { or } x \geqslant \frac{25}{3} .
\end{array}
$$

From (4) and (5), the solution to the inequality is
$$
x<\frac{5}{4} \sqrt{15} \text { or } x \geqslant \frac{25}{3} .
$$

If we use the idea of combining geometric and algebraic thinking to solve this inequality, it becomes much simpler.
Let $y^{2}=1$, then the given inequality becomes
$$
\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}>10 \text {. }
$$

This inequality indicates that the sum of the distances from the moving point $(x, y)$ to the points $F_{1}(-3,0)$ and $F_{2}(3,0)$ is greater than 10.
We notice that the equation
$$
\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}=10
$$

represents an ellipse with foci at $F_{1}(-3,0), F_{2}(3,0)$ and a semi-major axis of 5, whose equation is
$$
\frac{x^{2}}{25}+\frac{y^{2}}{16}=1
$$

Thus, this problem is equivalent to finding the x-coordinates of points on the line $y= \pm 1$ that lie outside the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, i.e., solving
$$
\left\{\begin{array}{l}
\frac{x^{2}}{25}+\frac{y^{2}}{16}>1 \\
y^{2}=1
\end{array}\right.
$$

From this, we get $x<\frac{5}{4} \sqrt{15}$ or $x \geqslant \frac{25}{3}$.