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.

A must be larger than $\frac {5\pi} 6 $ cos if it's not, then $\sin A\ge\frac 1 2$ so  $\sin B+\sin C\le\frac 1 2$ but $B+C\ge\frac {\pi} 6 $ which is not possible since $\sin B+\sin C>\sin(B+C)\ge\frac 1 2$
