Proof: Since $a^{2}-1$, $b^{2}-1$, and $c^{2}-1$ must have 2 that are not both greater than zero or not both less than zero, without loss of generality, let them be $a^{2}-1$ and $b^{2}-1$. Thus, we have $\left(a^{2}-1\right)\left(b^{2}-1\right) \leqslant 0$, which implies $a^{2}+b^{2} \geqslant a^{2} b^{2}+1$.
Therefore, $a^{2}+b^{2}+c^{2} \geqslant a^{2} b^{2}+1+c^{2}$, which means $a^{2} b^{2}+c^{2} \leqslant 2$.
Hence, the expressions $a^{2} b^{2}+c^{2}$, $b^{2} c^{2}+a^{2}$, and $c^{2} a^{2}+b^{2}$ must have at least one that is not greater than 2.