31 Find the smallest positive number $\lambda$, such that for any triangle with sides $a \geqslant \frac{b+c}{3}$, we have $a c + b c - c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right) \cdot(1993$ China National Training Team Test Question $)$
31. $a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c=(a+b-$ $c)^{2}+2 c(a+c-b)$

Let
$$I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{c(a+b-c)}=\frac{a+b-c}{2 c}+\frac{a+c-b}{a+b-c}$$

Since $a \geqslant \frac{b+c}{3}$, we have $a \geqslant \frac{1}{4}(a+b-c)+\frac{c}{2}$, thus, $a+c-b=2 a-(a+$ $b-c) \geqslant-\frac{1}{2}(a+b-c)+c$. Therefore,
$$\begin{aligned}
I \geqslant & \frac{a+b-c}{2 c}-\frac{\frac{1}{2}(a+b-c)}{a+b-c}+\frac{c}{a+b-c}= \\
& -\frac{1}{2}+\frac{a+b-c}{2 c}+\frac{c}{a+b-c} \geqslant \\
& -\frac{1}{2}+2 \sqrt{\frac{1}{2}}=\sqrt{2}-\frac{1}{2}
\end{aligned}$$

Thus,
$$\frac{a c+b c-c^{2}}{a^{2}+b^{2}+3 c^{2}+2 a b-4 b c}=\frac{1}{2 I} \leqslant \frac{1}{2 \sqrt{2}-1}=\frac{2 \sqrt{2}+1}{7}$$

Therefore, the required $\lambda \leqslant \frac{2 \sqrt{2}+1}{7}$.
On the other hand, when $a=\frac{\sqrt{2}}{4}+\frac{1}{2}, b=\frac{3 \sqrt{2}}{4}+\frac{1}{2}, c=$, then $a c+b c-c^{2}=\sqrt{2}, a^{2}+$ $b^{2}+3 c^{2}+2 a b-4 b c=4-\sqrt{2}$, so, $\frac{1}{2 I}=\frac{\sqrt{2}}{4-\sqrt{2}}=\frac{\sqrt{2}(4+\sqrt{2})}{14}=\frac{2 \sqrt{2}+1}{7}$, thus the required $\lambda \geqslant \frac{2 \sqrt{2}+1}{7}$.

In conclusion, $\lambda=\frac{2 \sqrt{2}+1}{7}$.