已知三角形$ABC$中，$A=\frac{ \pi }{3}$，D是$BC$边上一点，且满足$2BD=DC$，则$\frac{AD}{BC}$的最大值是．
思路一：
思路二：
∵$\vec{AD}=\frac{1}{3}\vec{AB}+\frac{2}{3}\vec{AC}$，${(\vec{AD})}^{2}={\left( \frac{1}{3}\vec{AB}+\frac{2}{3}\vec{AC} \right)}^{2}=\frac{1}{9}{c}^{2}+\frac{4}{9}{b}^{2}+\frac{2}{9}bc$．
由余弦定理得$B{C}^{2}={b}^{2}+{c}^{2}-bc$，则$\frac{{(AD)}^{2}}{{(BC)}^{2}}=\frac{{c}^{2}+4{b}^{2}+2bc}{9\left( {b}^{2}+{c}^{2}-bc \right)}$，
方法一：判别式法：令$y=\frac{{c}^{2}+4{b}^{2}+2bc}{{b}^{2}+{c}^{2}-bc}$，$(4-y){b}^{2}+(2+y)bc+(1-y){c}^{2}=0$有解，
$\mathrm{~\Delta~}={(2+y)}^{2}-4(4-y)(1-y)=-3{y}^{2}+24y-12 \geq 0$，解得$4-2\sqrt{3} \leq y \leq 4+2\sqrt{3}$
$\frac{{(AD)}^{2}}{{(BC)}^{2}}=\frac{1}{9}y$．∴$\frac{AD}{BC} \leq \frac{\sqrt{3}+1}{3}$
方法二：换元法$\frac{{(AD)}^{2}}{{(BC)}^{2}}=\frac{{c}^{2}+4{b}^{2}+2bc}{9\left( {b}^{2}+{c}^{2}-bc \right)}=\frac{1}{9}\frac{1+4{\left( \frac{b}{c} \right)}^{2}+2\frac{b}{c}}{1+{\left( \frac{b}{c} \right)}^{2}-\frac{b}{c}}$．
令$\frac{b}{c}=t$
上式$=\frac{4{t}^{2}+2t+1}{9\left( {t}^{2}-t+1 \right)}=\frac{4\left( {t}^{2}-t+1 \right)+6t-3}{9\left( {t}^{2}-t+1 \right)}=\frac{4}{9}+\frac{2t-1}{3\left( {t}^{2}-t+1 \right)}$
令$2t-1=m$，则有$\frac{2t-1}{\left( {t}^{2}-t+1 \right)}=\frac{4m}{{m}^{2}+3}=\frac{4}{m+\frac{3}{m}} \leq \frac{2\sqrt{3}}{3}$，
$ \therefore \frac{{(AD)}^{2}}{{(BC)}^{2}} \leq \frac{4+2\sqrt{3}}{9}$，∴$\frac{AD}{BC} \leq \frac{\sqrt{3}+1}{3}$
故答案为：$\frac{\sqrt{3}+1}{3}$