Example 10 Find the smallest positive number $\lambda$, such that for any triangle with side lengths $a, b, c$, as long as $a \geqslant \frac{b+c}{3}$, we have
$$
a c+b c-c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right) .
$$
(1993 Chinese National Team Test Question)
Solving, we know
$$
\begin{array}{l} 
a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c \\
=(a+b-c)^{2}+2 c(a+c-b) . \\
\text { Let } I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{2 c(a+b-c)}=\frac{a+b-c}{2 c}+\frac{a+c-b}{a+b-c} .
\end{array}
$$

Since $a \geqslant \frac{1}{3}(b+c)$, 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, $I \geqslant \frac{a+b-c}{2 c}-\frac{\frac{1}{2}(a+b-c)}{a+b-c}+\frac{c}{a+b-c}$
$$
=-\frac{1}{2}+\frac{a+b-c}{2 c}+\frac{c}{a+b-c} \geqslant-\frac{1}{2}+2 \sqrt{\frac{1}{2}}=\sqrt{2}-\frac{1}{2} \text {, }
$$

which means $\frac{a c+b c-c^{2}}{a^{2}+b^{2}+3 c^{2}+2 a b-4 b c}$. Thus, the required $\lambda \leqslant \frac{2 \sqrt{2}+1}{7}$.
On the other hand, when $a=\frac{\sqrt{2}}{4}+\frac{1}{2}, b=\frac{3 \sqrt{2}}{4}+\frac{1}{2}, c=1$, then
$$
a c+b c-c^{2}=\sqrt{2}, a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=4-\sqrt{2} \text {, }
$$

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