Proof: Let
$$f(a, b, c, d)=\frac{1}{(3 a-1)^{2}}+\frac{1}{(3 b-1)^{2}}+\frac{1}{(3 c-1)^{2}}+\frac{1}{(3 d-1)^{2}}$$

If $\min \{a, b, c, d\}>\frac{1}{3}$, then by the AM - GM inequality we have
$$\begin{aligned} f(a, b, c, d)-f(\sqrt{a d}, b, c, \sqrt{a d}) \geqslant & \frac{2}{(3 a-1)(3 d-1)}-\frac{2}{(3 \sqrt{a d}-1)^{2}}= \\ & \frac{2(3 a+3 d-6 \sqrt{a d})}{(3 a-1)(3 d-1)(3 \sqrt{a d}-1)^{2}} \geqslant 0 \end{aligned}$$

Similarly,
$$f(\sqrt{a d}, b, c, \sqrt{a d}) \geqslant f(\sqrt{a d}, \sqrt{b c}, \sqrt{b c}, \sqrt{a d})$$

Let $\sqrt{a d}=x$, then $\sqrt{b c}=\frac{1}{x}$, and using the AM - GM inequality again we have
$$\begin{array}{l} f(\sqrt{a d}, \sqrt{b c}, \sqrt{b c}, \sqrt{a d})=\frac{2}{(3 x-1)^{2}}+\frac{2}{\left(\frac{3}{x}-1\right)^{2}} \geqslant \\ \frac{4}{(3 x-1)\left(\frac{3}{x}-1\right)}=\frac{4}{10-3 x-\frac{3}{x}} \geqslant 1 \end{array}$$

Thus the proposition is proved.
If $\min \{a, b, c, d\}<\frac{1}{3}$, without loss of generality, let $a=\min \{a, b, c, d\}$.
Notice that
$$\frac{1}{(3 a-1)^{2}} \geqslant 1 \Leftrightarrow a(9 a-6) \leqslant 0 \Leftrightarrow 0 \leqslant a \leqslant \frac{2}{3}$$

That is, the inequality also holds in this case, so the original inequality is proved.
Sometimes it is not necessary to adjust the variables to be equal or to zero; the adjustment target may be the average of all variables.