Minimum value of angles of a triangle In a triangle $ABC$, if $\sin A+\sin B+\sin C\leq1$,then prove that $$\min(A+B,B+C,C+A)<\pi/6$$
where $A,B,C$ are angles of the triangle in radians.
if we assume $A>B>C$,then $\sum \sin A\leq 3 \sin A$,and $ A\geq \frac{A+B+C}{3}=\pi/3$.also $\sum \sin A\geq 3\sin C$ and  $ C\leq \frac{A+B+C}{3}=\pi/3$.But I could not proceed with this. Please help me in this regard.Thanks.

Since you assumed $A\geq B\geq C$, it must be that $\dfrac{A}{2}+C\leq\dfrac{\pi}{2}.$ Hence, $\sin\tfrac{A}{2}<\sin(\tfrac{A}{2}+C) = \cos(\tfrac{B-C}{2})$. 
Finally, $$1\geq \sin A+\sin B+\sin C = \sin A+2\sin\tfrac{B+C}{2}\cos\tfrac{B-C}{2} = 2\cos\tfrac{A}{2}\big(\sin\tfrac{A}{2}+\cos\tfrac{B-C}{2}\big)>4\cos\tfrac{A}{2}\sin\tfrac{A}{2} = 2\sin A\Rightarrow$$
$\sin A< \tfrac{1}{2}$, which means that $A\gt\frac{5\pi}{6}$, because $A$ is the largest angle, by assumption. Then, $B+C<\tfrac{\pi}{6}$ and we are done. 
