6. Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ be the three interior angles of $\triangle \mathrm{ABC}$, then the maximum value of the real number $\lambda$ such that $\frac{1}{\sin A}+\frac{1}{\sin B} \geq \frac{\lambda}{3+2 \cos C}$ always holds is $\qquad$ ;
Solution: Since $3+2 \cos C=4 \cos ^{2} \frac{C}{2}+1>0$, therefore, $\lambda \leq\left(4 \cos ^{2} \frac{C}{2}+1\right)\left(\frac{1}{\sin A}+\frac{1}{\sin B}\right)$,
Since $\left(4 \cos ^{2} \frac{C}{2}+1\right)\left(\frac{1}{\sin A}+\frac{1}{\sin B}\right) \geq \frac{8 \cos \frac{C}{2}}{\sqrt{\sin A \sin B}}=\frac{8 \cos \frac{C}{2}}{\sqrt{\frac{1}{2}[\cos (A-B)+\cos C]}}$
$\geq \frac{8 \cos \frac{C}{2}}{\sqrt{\frac{1}{2}(1+\cos C)}}=\frac{8 \cos \frac{C}{2}}{\cos \frac{C}{2}}=8$, equality holds only when $\mathrm{A}=\mathrm{B}=30^{\circ}, \mathrm{C}=120^{\circ}$, so $\lambda_{\text {maximum }}=8$.