The bisectors of triangle ABC meet its circumcircle in points D,E,F respectively show that the area of Triangle ABC cannot be greater then that of triangle DEF
The area of triangle ABC is , where R is the radius of the circumcircle. Since triangle DEF has the same circumcircle as ABC, its area is . So we want to show that . But the angles D, E, F are equal to (as you can see by drawing a picture and looking for angles in the same segment). So we want to prove that .
Since all these sines are positive, this is equivalent to the inequality . But we can split that up as the product of three inequalities (and two other, similar, inequalities).
I'm sure there are several ways to prove that last inequality. One way would be to take logs and use the fact that the function is concave for . But I expect that there is also a purely trigonometric proof.
We can obtain the vertex angles for triangle as shown in the attached images.
In Fig.2, using the common chord
Using the common chord
Similarly for the other angles in Fig.3 and Fig.4.
The angles of triangle are therefore
while the angles of triangle are
The question becomes
This inequality holds if the following 3 inequalities hold
Archie Meade's comment shows how I could have completed the proof of this result. In comment #3 above, I showed how the proof could be reduced to proving inequalities of the form . If we let and then that inequality becomes . But that is equivalent to , which is obviously true.