wlogA>B>C, the problem is equivalent to sin(A/2)sin(B/2)sin(C/2)<=sin((B+C)/4)sin((C+A)/4)sin((A+B)/4),use the Karamata Inequality with lnsin(x),done

Anant Mudgal (IND 2) gave a synthetic solution to this problem, for which he won a special prize.

Ok, the special prize might have been given because no one even tried synthetic; and gave an algebraic proof mostly. We first recall the Euler Poncelet Formula, $d=\sqrt{R^2-2Rr}$ where the notations are what they usually mean. Note also that the incenter of $ABC$ is the DeLongChamps point of triangle $A_1B_1C_1$. We also see that triangle $A_1B_1C_1$ is acute angled, being similar to the intouch triangle of $ABC$. Now the result which we seek to prove is the following: In acute angled triangle $XYZ$ with orthocenter $H$, incenter $I$ and circumcenter $O$, we have $OH \ge OI$. The proof is as follows: Assume wlog that $\angle X$ is the median of the set $\{\angle X,\angle Y,\angle Z\}$. Since $H,O$ are isogonal conjugates, we see that $YI$ bisects angle $OYH$ and $ZI$ bisects angle $OZH$. Let $YH \cap ZO=P$ and $YO \cap ZH=Q$. We claim that $HPOQ$ is a convex quadrilateral and $I$ lies in its interior. We can see that $YHOZ$ is a convex quadrilateral by our angle condition on $X$. Therefore, $P,Q$ lies on opposite sides of $HO$ and hence $HPOQ$ is convex. Now, $I$ lies in its interior immediately follows. Observe that $\angle OPH=180^{\circ}-\angle Y >90^{\circ}$ and $\angle OQH=180^{\circ}-\angle Z>90^{\circ}$. Let $OI$ extended meet the boundary of the quadrilateral $OPHQ$ at point $T$. We clearly have $\angle OTH>\min\{\angle OPH,\angle OQH\}>90^{\circ}$. Therefore, $OI \le OT \le OH$. The result holds. Comment This was the only time in India TSTs when a solution graded $9/10$ (due to some technical issue) got a special prize.

This might be easy but somebody please give a complete trigonometric solution.

PRO2000 wrote: This might be easy but somebody please give a complete trigonometric solution. I think the inequality $sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2}) \leq sin(\frac{A+B}{4})sin(\frac{A+C}{4})sin(\frac{C+B}{4})$ Can be reached easily by A.M-G.M. on $sin(\frac{A}{4})cos(\frac{B}{4})$ and $sin(\frac{B}{4})cos(\frac{A}{4})$ Which gives $sin(\frac{A+B}{4}) \geq 2(sin(\frac{A}{4})cos(\frac{B}{4})sin(\frac{B}{4})cos(\frac{A}{4}))^\frac{1}{2}$ Now get the other two similar inequalities and multiply them to get the answer.

Solution: We know that $\sum_\text{cyc} \cos A = 1 + \frac{r}{R}$, so we just need that $\sum_\text{cyc} \cos A \le \sum_\text{cyc} \sin \frac A2$. But this is easy because $\sum_\text{cyc} \cos A = \sum_\text{cyc} \frac{\cos B + \cos C}{2} = \sum_\text{cyc} \cos \left( \frac{B-C}{2} \right) \sin \frac A2 \le \sum_\text{cyc} \sin \frac A2$

From the IMOSL 1988 problem here (and my solution using Schur's ), we know that the area of $A_1 B_1 C_1$ is greater than or equal to that of $ABC$. It's quite easy to see that the semi-perimeters of the two triangles is the same. Since $r = \frac{\Delta}{s}$, the conclusion follows immediately. EDIT: As @below highlights, this is not valid. Basically what I was assuming is if an arc doubles in length, so does its associated chord. This is, of course, false

Arthur. wrote: It's quite easy to see that the semi-perimeters of the two triangles is the same. I don't think so.

anantmudgal09 wrote: Ok, the special prize might have been given because no one even tried synthetic; and gave an algebraic proof mostly. We first recall the Euler Poncelet Formula, $d=\sqrt{R^2-2Rr}$ where the notations are what they usually mean. Note also that the incenter of $ABC$ is the DeLongChamps point of triangle $A_1B_1C_1$. We also see that triangle $A_1B_1C_1$ is acute angled, being similar to the intouch triangle of $ABC$. Now the result which we seek to prove is the following: In acute angled triangle $XYZ$ with orthocenter $H$, incenter $I$ and circumcenter $O$, we have $OH \ge OI$. The proof is as follows: Assume wlog that $\angle X$ is the median of the set $\{\angle X,\angle Y,\angle Z\}$. Since $H,O$ are isogonal conjugates, we see that $YI$ bisects angle $OYH$ and $ZI$ bisects angle $OZH$. Let $YH \cap ZO=P$ and $YO \cap ZH=Q$. We claim that $HPOQ$ is a convex quadrilateral and $I$ lies in its interior. We can see that $YHOZ$ is a convex quadrilateral by our angle condition on $X$. Therefore, $P,Q$ lies on opposite sides of $HO$ and hence $HPOQ$ is convex. Now, $I$ lies in its interior immediately follows. Observe that $\angle OPH=180^{\circ}-\angle Y >90^{\circ}$ and $\angle OQH=180^{\circ}-\angle Z>90^{\circ}$. Let $OI$ extended meet the boundary of the quadrilateral $OPHQ$ at point $T$. We clearly have $\angle OTH>\min\{\angle OPH,\angle OQH\}>90^{\circ}$. Therefore, $OI \le OT \le OH$. The result holds. Comment This was the only time in India TSTs when a solution graded $9/10$ (due to some technical issue) got a special prize. What is the name of the prize????

Ehh,complex bashing can be used,as mentioned in egmo.