math contest geometry problem Consider a triangle $ABC$ with circumcircle $\omega$. Let $O$ be the center of $\omega$ and let $D, E, F$ be the
midpoints of minor arcs $BC, CA, AB$ respectively. Let $DO$ intersect $\omega$ again at a point $A'$.
Define $B'$ and $C'$ similarly. Prove that 
$ [ABC] \leq [A'B'C'] $ 
$[X]$ denotes area of $X$.
 A: Let $\angle AOB=2\gamma$, $\angle BOC=2\alpha$, $\angle COA=2\beta$, so that:
$$
[ABC]={1\over2}r^2(\sin2\alpha+\sin2\beta+\sin2\gamma),
$$
where $r$ is the radius of $\omega$.
On the other hand it follows from the definition of $DEF$ that
$$
[DEF]={1\over2}r^2(\sin(\alpha+\beta)+\sin(\beta+\gamma)+\sin(\gamma+\alpha)).
$$
As $[A'B'C']=[DEF]$, we must prove $[ABC]\le[DEF]$, that is
$$
\sin2\alpha+\sin2\beta+\sin2\gamma\le \sin(\alpha+\beta)+\sin(\beta+\gamma)+\sin(\gamma+\alpha).
$$
As $\alpha+\beta+\gamma=\pi$ we can eliminate $\gamma$ from the above inequality, which then becomes:
$$
\sin2\alpha+\sin2\beta-\sin2(\alpha+\beta)\le \sin(\alpha+\beta)+\sin\alpha+\sin\beta.
$$
This can also be put in the form $F(\alpha,\beta)\ge0$, where:
$$
F(\alpha,\beta)=\sin\alpha+\sin\beta
+\sin(\alpha+\beta)(1-4\sin\alpha\sin\beta)
$$
This inequality must be proved for $0\le\alpha,\beta\le\pi/3$, because we can take as $\gamma$ the largest among the angles. The proof can be then carried out with the standard techniques of calculus, to show that the minimum of $F$ is zero.
