Proof: Let $\frac{a}{b}=x, \frac{b}{c}=y, \frac{c}{a}=z$, then the original inequality is equivalent to, if positive numbers $x, y, z$ satisfy $x y z=1$, then the inequality $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+3 \geqslant 2(x+y+z)$ holds, which is $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+3 \geqslant 2(x+y+z)$. Noting that $x^{2} y^{2} z^{2}=1$, it follows that at least two of $x^{2}, y^{2}, z^{2}$ are simultaneously not less than 1, or not greater than 1. Without loss of generality, assume they are $x^{2}$ and $y^{2}$, then $\left(x^{2}-1\right)\left(y^{2}-1\right) \geqslant 0$.

Below, we use the method of difference to prove inequality (*). In fact,
$x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+3-2(x+y+z)$
$=x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+3-2(x+y+z)$
$=\left(y^{2} z^{2}+z^{2} x^{2}-2 y z \cdot z x\right)+\left(x^{2} y^{2}-x^{2}-y^{2}\right.$
$+1)+\left(x^{2}-2 x+1\right)+\left(y^{2}-2 y+1\right)$
$=(y z-z x)^{2}+\left(x^{2}-1\right)\left(y^{2}-1\right)+(x-1)^{2}+$ $(y-1)^{2} \geqslant 0$.
This shows that inequality (*) holds, hence the original inequality is proved.