hajimbrak wrote: Prove that for any triangle $ABC$, the inequality $\displaystyle\sum_{\text{cyclic}}\cos A\le\sum_{\text{cyclic}}\sin (A/2)$ holds. $\displaystyle\sum_{\text{cyclic}}\cos A=\sum_{\text{cyclic}}\sin (A/2)\cos{\frac{B-C}{2}}$

http://www.artofproblemsolving.com/community/c6h622489p3722650

Prove that for any triangle $ABC$, the inequality $\displaystyle\sum_{\text{cyclic}}\sin A\le\sum_{\text{cyclic}}\cos \frac{A}{2}$ holds.

sqing wrote: Prove that for any triangle $ABC$, the inequality $\displaystyle\sum_{\text{cyclic}}\sin A\le\sum_{\text{cyclic}}\cos \frac{A}{2}$ holds. $\sin A+\sin B=2\cos{\frac{C}{2}}\cos\frac{A-B}{2}\le 2\cos\frac{C}{2}$ Similarly，we have two another concyclic inequalities,add them and we have proved the conclusion It's a very beautiful conclusion.

$\displaystyle\sum_{\text{cyclic}}\cos A=1+ 4sin A/2sinB/2sinC/2$ and $\displaystyle\sum_{\text{cyclic}}\sin (A/2)= 1+4sin(B+C/2)sin(C+A/2)sin(A+B/2)$ which reduces to prove $sin(A/2)sin(B/2)sin(C/2)\le sin(B+C/2)sin(A+B/2)sin(C+A/2)$ which is trivial.

I think this is the easiest imotc problem ever. Did it in 5 min

Wlog take A>60 CosA<=sin(A/2) reduces (sinA/2 +1)(2sinA/2–1)>0 Since 0<A/2<π SinA/2>0 And because of our assumption 2nd part is >0

This is such a JEE-type problem- how did it sneak its way into $TC$?

In $ \triangle ABC .$ Prove that $$\sin A\sin B\sin C\le\frac{3\sqrt3}{8}\sin(A+\frac{B}{2})\sin(B+\frac{C}{2})\sin(C+\frac{A}{2})$$p/9076044578

Mate007 wrote: $\displaystyle\sum_{\text{cyclic}}\cos A=1+ 4sin A/2sinB/2sinC/2$ and $\displaystyle\sum_{\text{cyclic}}\sin (A/2)= 1+4sin(B+C/2)sin(C+A/2)sin(A+B/2)$ which reduces to prove $sin(A/2)sin(B/2)sin(C/2)\le sin(B+C/2)sin(A+B/2)sin(C+A/2)$ which is trivial. Excellent, of course