Example 14 Find the minimum value of the function $f(x)=\sqrt{x^{2}+4}+\sqrt{x^{2}+2 x+10}$.
Solution 1 The domain of the function $f(x)$ is $(-\infty,+\infty)$, and by the given condition, $\sqrt{x^{2}+4}, \frac{1}{2} f(x)$, $\sqrt{x^{2}+2 x+10}$ form an arithmetic sequence, so we can let $\sqrt{x^{2}+4}=\frac{1}{2} f(x)-d$,
$$
\sqrt{x^{2}+2 x+10}=\frac{1}{2} f(x)+d,
$$

Since $\sqrt{x^{2}+4} \geqslant 2, \sqrt{x^{2}+2 x+10}=\sqrt{(x+1)^{2}+9} \geqslant 3$, thus $y \geqslant 5$. By squaring (2) and subtracting (1) squared, we get $x=d \cdot f(x)-3$.
Substituting (3) into (1), we get $4\left[f^{2}(x)-1\right] d^{2}+20 \cdot f(x) \cdot d+52-f^{2}(x)=0$.
By $\Delta=[20 \cdot f(x)]^{2}-4 \cdot 4\left[f^{2}(x)-1\right] \cdot\left[52-f^{2}(x)\right] \geqslant 0$, we get $f^{2}(x) \geqslant 26$ or $f^{2}(x) \leqslant 2$ (discard).
Therefore, $f(x) \geqslant \sqrt{26}$. When $f(x)=26$, $d=\frac{\sqrt{26}}{10}, x=-\frac{2}{5} \in \mathbf{R}$, which meets the condition, so $f(x)_{\min }=\sqrt{26}$.
Solution 2 To find the minimum value of $\sqrt{x^{2}+2^{2}}+\sqrt{(x+1)^{2}+3^{2}}$, since $x \in$ $(-\infty,+\infty)$, we know $-1<x<0$, thus $x+1=1-|x|$. Construct a right triangle $\triangle A B C$ such that $\angle B=90^{\circ}, A B=2, B C=|x|$, and construct a right triangle $\triangle C D E$ such that $\angle D=90^{\circ}, C D=1-|x|, D E=3$, as shown in Figure 5-7.
Hence $\sqrt{x^{2}+4}+\sqrt{x^{2}+2 x+10}=A C+C E \geqslant A E$.
And $A E=\sqrt{B D^{2}+(A B+D E)^{2}}=\sqrt{1+5^{2}}=\sqrt{26}$.
The equality holds when $\mathrm{Rt} \triangle A B C \sim \mathrm{Rt} \triangle E D C$, solving for $|x|=\frac{2}{5}$, i.e., $x=-\frac{2}{5}$.