Let $S=\frac{1}{a^{2}}-a^{2}+\frac{1}{b^{2}}-b^{2}+\frac{1}{c^{2}}-c^{2}$. If we can find that the minimum value of $S$ is 0, then the conclusion is proven.
Let $a \leqslant b \leqslant c$, and for any given $c=c_{0}\left(1 \leqslant c_{0} \leqslant 3\right)$. Then $b+a=3-c_{0}$.
Let $x=b-a \geqslant 0$. Then
$b=\frac{3-c_{0}+x}{2}, a=\frac{3-c_{0}-x}{2}$.
Substituting into $S$ we get
$$\begin{aligned} S= & \frac{4}{\left(3-c_{0}-x\right)^{2}}-\frac{\left(3-c_{0}-x\right)^{2}}{4}+\frac{4}{\left(3-c_{0}+x\right)^{2}}- \\ & \frac{\left(3-c_{0}+x\right)^{2}}{4}+\frac{1}{c_{0}^{2}}-c_{0}^{2} . \end{aligned}$$

Differentiating with respect to $x$ we get
$$\begin{aligned} S^{\prime}(x) & =\frac{-2 \times 4(-1)}{\left(3-c_{0}-x\right)^{3}}-\frac{1}{4} \times 2\left(3-c_{0}-x\right)(-1)+ \\ & 4(-2) \frac{1}{\left(3-c_{0}+x\right)^{3}}-\frac{1}{4} \times 2\left(3-c_{0}+x\right) \\ = & \frac{8}{\left(3-c_{0}-x\right)^{3}}-\frac{8}{\left(3-c_{0}+x\right)^{3}}-x \\ = & \frac{1}{a^{3}}-\frac{1}{b^{3}}-(b-a) \\ = & (b-a) \frac{a^{2}+a b+b^{2}-a^{3} b^{3}}{a^{3} b^{3}} . \end{aligned}$$

Since $c_{0} \geqslant 1$, we have
$a+b \leqslant 2 \Rightarrow a b \leqslant\left(\frac{a+b}{2}\right)^{2} \leqslant 1$.
Thus, $a b-a^{3} b^{3} \geqslant 0$, and $b-a \geqslant 0$.
Therefore, $S^{\prime}(x) \geqslant 0$.
$S$ is an increasing function of $x$, so for a given $c_{0}$, the minimum value of $S$ is always achieved when $x=0\left(b=a=\frac{3-c_{0}}{2}\right)$.
Substituting into equation (1) we get
$$S=c_{0}^{-2}-c_{0}^{2}+8\left(3-c_{0}\right)^{-2}-\frac{1}{2}\left(3-c_{0}\right)^{2} \text {. }$$

Differentiating with respect to $c_{0}$ we get
$$\begin{aligned} S^{\prime}\left(c_{0}\right)= & -2 c_{0}^{-3}-2 c_{0}+8(-2)\left(3-c_{0}\right)^{-3}(-1)- \\ & \frac{1}{2} \times 2\left(3-c_{0}\right)(-1) \end{aligned}$$
$$\begin{array}{l} =2\left[\frac{1}{\left(\frac{3-c_{0}}{2}\right)^{3}}-\frac{1}{c_{0}^{3}}\right]+3-3 c_{0} \\ =\left(3 c_{0}-3\right) \frac{c_{0}^{2}+c_{0} \cdot \frac{3-c_{0}}{2}+\left(\frac{3-c_{0}}{2}\right)^{2}-c_{0}^{3}\left(\frac{3-c_{0}}{2}\right)^{3}}{c_{0}^{3}\left(\frac{3-c_{0}}{2}\right)^{3}} . \end{array}$$

It is easy to prove: $1-c_{0}^{2}\left(\frac{3-c_{0}}{2}\right)^{2} \geqslant-2\left(\Leftrightarrow c_{0} \cdot \frac{3-c_{0}}{2}\right.$ $\leqslant \sqrt{3}$. And $c_{0} \cdot \frac{3-c_{0}}{2}=2 \cdot \frac{c_{0}}{2}\left(\frac{3}{2}-\frac{c_{0}}{2}\right) \leqslant 2\left(\frac{3}{4}\right)^{2}$ $=\frac{9}{8}<\sqrt{3}$ holds).
Thus, $S^{\prime}\left(c_{0}\right)$
$$\begin{array}{l} \geqslant\left(3 c_{0}-3\right) \frac{c_{0}^{2}-2 c_{0} \cdot \frac{3-c_{0}}{2}+\left(\frac{3-c_{0}}{2}\right)^{2}}{c_{0}^{3}\left(\frac{3-c_{0}}{2}\right)^{3}} \\ =\left(3 c_{0}-3\right) \frac{\left(c_{0}-\frac{3-c_{0}}{2}\right)^{2}}{c_{0}^{3} \cdot\left(\frac{3-c_{0}}{2}\right)^{3}} \geqslant 0 . \end{array}$$

This shows that $S$ is an increasing function of $c_{0}$.
Since $1 \leqslant c_{0} \leqslant 3$, when $c_{0}=1$ (i.e., $b=a=\frac{3-1}{2}=1$), $S_{\text {min }}=\frac{1}{1^{2}}-1^{2}+\frac{1}{1^{2}}-1^{2}+\frac{1}{1^{2}}-1^{2}=0$. Therefore, $S \geqslant 0$, and the original statement is proven.