66. Let the sides of $\triangle ABC$ be $a, b, c$, and the corresponding medians be $m_{a}, m_{b}, m_{c}$, and the angle bisectors be $w_{a}, w_{b}, w_{c}$. Suppose $w_{a} \cap m_{b}=P, w_{b} \cap m_{c}=Q, w_{c} \cap m_{a}=R$. Let the area of $\triangle PQR$ be $\delta$, and the area of $\triangle ABC$ be $F$. Find the smallest positive constant $\lambda$ such that the inequality $\frac{\delta}{F}<\lambda$ holds. (2003 China National Training Team)
66. Let $a \geqslant b \geqslant c$, and let $G$ be the centroid of $\triangle ABC$. Then it is easy to see that $P$ lies on $BG$. Let $D$ be the midpoint of $BC$, and let $AP$ intersect $BC$ at $E$. By the Angle Bisector Theorem, we have $\frac{BE}{ED}=\frac{2c}{b-c}$.

Applying Menelaus' Theorem to $\triangle GBD$ and the transversal $APE$, we get
$$\frac{GP}{PB} \cdot \frac{BE}{ED} \cdot \frac{DA}{AG}=1 \text{, hence } \frac{GP}{GB}=\frac{b-c}{b+2c} \text{.}$$

Similarly, we can obtain
$$\begin{array}{l}
\frac{S_{\triangle GRQ}}{F}=\frac{1}{3} \cdot \frac{(a-b)(a-c)}{(a+2b)(c+2a)} \\
\frac{S_{\triangle GPR}}{F}=\frac{1}{3} \cdot \frac{(a-b)(b-c)}{(a+2b)(b+2c)}
\end{array}$$

Therefore,
$$\begin{array}{l}
\frac{\delta}{F}=\frac{S_{\triangle GPQ}+S_{\triangle GRQ}-S_{\triangle GPR}}{F}= \\
\frac{ab^2+bc^2+ca^2-3abc}{(a+2b)(b+2c)(c+2a)}
\end{array}$$

Let
$$f(a, b, c)=\frac{ab^2+bc^2+ca^2-3abc}{(a+2b)(b+2c)(c+2a)}$$

Now let $a=b=1, c \rightarrow 0$, then $f(a, b, c) \rightarrow \frac{1}{6}$.
We need to prove
$$f(a, b, c)<\frac{1}{6}$$

Through constant transformation, it is easy to see that (1) is equivalent to $2|(a-b)(b-c)(c-a)|<27abc$, which is clearly true, because $2|(a-b)(b-c)(c-a)|<2abc<27abc$.

In conclusion, the smallest positive constant is $\frac{1}{6}$.