2 Let the medians of $\triangle A B C$ corresponding to sides $a, b, c$ 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 $\triangle P Q R$ be $\delta$, and the area of $\triangle A B C$ be $F$. Find the smallest positive constant $\lambda$ such that the inequality $\frac{\delta}{F}<\lambda$ holds.
Let's assume $a \geqslant b \geqslant c$, and let $G$ be the centroid of $\triangle ABC$. It is easy to see that $P$ lies on $BG$, $R$ lies on $AG$, and $Q$ lies on the extension of $CG$. 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 $\frac{GP}{PB} \cdot \frac{BE}{ED} \cdot \frac{DA}{AG} = 1$, hence $\frac{GP}{GB} = \frac{b-c}{b+2c}$. Similarly, $\frac{GQ}{GC} = \frac{a-c}{c+2a}$. Noting the area relationship $F = 3 S_{\triangle GBC}$, we have
$$
\frac{S_{\triangle GPQ}}{F} = \frac{1}{3} \cdot \frac{GP \cdot GQ}{GB \cdot GC} = \frac{1}{3} \cdot \frac{(b-c)(a-c)}{(b+2c)(c+2a)}.
$$

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

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

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 we get $f(a, b, c) \rightarrow \frac{1}{6}$.
Next, we prove
$$
f(a, b, c) < \frac{1}{6}.
$$

Through constant transformation, it is easy to see that (*) is equivalent to $2|(a-b)(b-c)(c-a)| < 27abc$, which is clearly true (since $2|(a-b)(b-c)(c-a)| < 2cab < 27abc$).
In conclusion, the smallest positive constant is $\frac{1}{6}$.