6. Find the smallest number $k$ such that $\frac{t_{a}+t_{b}}{a+b}<k$, where $a$ and $b$ are two sides of a triangle, and $t_{a}$ and $t_{b}$ are the lengths of the angle bisectors corresponding to these sides.
(2002, Bulgaria National Mathematical Olympiad (Final))
When $\triangle A B C$ satisfies $a=b$, and the base angle $\angle A \rightarrow 0$,
$$
\begin{array}{l}
c \rightarrow 2 b, \cos \frac{A}{2} \rightarrow 1, \\
t_{a}=\frac{b c \sin A}{(b+c) \sin \frac{A}{2}}=\frac{2 b c}{b+c} \cos \frac{A}{2} .
\end{array}
$$

Thus, $\frac{t_{a}+t_{b}}{a+b}=\frac{t_{a}}{a} \rightarrow \frac{4}{3}$. Therefore, $k \geqslant \frac{4}{3}$.
$$
\text { Also } t_{a}2\left(\frac{2}{3}-\frac{b}{a+2 b}-\frac{a}{2 a+b}\right) \\
=\frac{2(a-b)^{2}}{3(a+2 b)(2 a+b)} \geq 0 .
\end{array}
$$

Hence $\frac{t_{a}+t_{b}}{a+b}<\frac{4}{3}$. Therefore, the minimum value of $k$ is $\frac{4}{3}$.