Example 2.3.5 (Bulgarian MO 2002) Find the smallest number $k$, such that $\frac{t_{a}+t_{b}}{a+b}<k$, where $a, b$ are two sides of a triangle, and $t_{a}, t_{b}$ are the lengths of the angle bisectors corresponding to these two sides.
Prove by first considering the extreme case: $\triangle A B C$ satisfies $a=b$, and the base angle $\angle A \rightarrow 0$ when $c \rightarrow 2 b$, and 
\[ c2\left(\frac{2}{3}-\frac{b}{a+2 b}-\frac{a}{2 a+b}\right)=\frac{2(a-b)^{2}}{3(a+2 b)(2 a+b)} \geqslant 0 . \]

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