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 $\displaystyle 2R^2\sin A\sin B\sin C$, where R is the radius of the circumcircle. Since triangle DEF has the same circumcircle as ABC, its area is $\displaystyle 2R^2\sin D\sin E\sin F$. So we want to show that $\displaystyle \sin D\sin E\sin F \geqslant \sin A\sin B\sin C$. But the angles D, E, F are equal to $\displaystyle \frac{B+C}2,\ \frac{C+A}2,\ \frac{A+B}2$ (as you can see by drawing a picture and looking for angles in the same segment). So we want to prove that $\displaystyle \sin \frac{B+C}2\sin \frac{C+A}2\sin\frac{A+B}2 \geqslant \sin A\sin B\sin C$.
Since all these sines are positive, this is equivalent to the inequality $\displaystyle \sin^2 \frac{B+C}2\sin^2 \frac{C+A}2\sin^2\frac{A+B}2 \geqslant \sin^2 A\sin^2 B\sin^2 C$. But we can split that up as the product of three inequalities $\displaystyle \sin^2 \frac{B+C}2\geqslant \sin B\sin C$ (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 $\displaystyle \log\sin x$ is concave for $\displaystyle 0\leqslant x\leqslant \pi$. But I expect that there is also a purely trigonometric proof.
We can obtain the vertex angles for triangle $\displaystyle D$$\displaystyle E$$\displaystyle F$ as shown in the attached images.
In Fig.2, using the common chord $\displaystyle BE,\;\;|\angle\ BCE|=|\angle\ BFE|=Z$
Using the common chord $\displaystyle CF,\;\;|\angle\ CBF|=|\angle\ CEF|=Y$
Similarly for the other angles in Fig.3 and Fig.4.
The angles of triangle $\displaystyle D$$\displaystyle E$$\displaystyle F$ are therefore $\displaystyle Y+Z,\;\;X+Y,\;\;X+Z$
while the angles of triangle $\displaystyle ABC$ are $\displaystyle 2X,\;\;2Y,\;\;2Z.$
$\displaystyle R=circumcircle\ radius$
$\displaystyle Area\ \triangle\ ABC=2R^2sin2Xsin2Ysin2Z$
$\displaystyle Area\ \triangle\ $$\displaystyle D$$\displaystyle E$$\displaystyle F$$\displaystyle =2R^2sin(Y+Z)sin(X+Y)sin(X+Z)$
The question becomes
$\displaystyle (sin2X)(sin2Y)(sin2Z)\ \le\ sin(Y+Z)sin(X+Y)sin(X+Z)\;\;?$
$\displaystyle (2sinXcosX)(2sinYcosY)(2sinZcosZ)$
$\displaystyle \le\ (sinYcosZ+sinZcosY)(sinXcosY+sinYcosX)(sinXcosZ+si nZcosX)\;\;?$
$\displaystyle 8sinXsinYsinZcosXcosYcosZ$
$\displaystyle \le\ (sinXsinYcosYcosZ+sin^2YcosXcosZ+)(sinXcosZ+sinZco sX)$
$\displaystyle +(sinXsinZcos^2Y+sinYsinZcosXcosY)(sinXcosZ+sinZco sX)\;\;?$
$\displaystyle 8sinXsinYsinZcosXcosYcosZ$
$\displaystyle \le\ sin^2XsinYcosYcos^2Z+sinXsinYsinZcosXcosYcosZ$
$\displaystyle +sinXsin^2YcosXcos^2Z+sin^2YsinZcos^2XcosZ$
$\displaystyle +sin^2XsinZcos^2YcosZ+sinXsin^2ZcosXcos^2Y$
$\displaystyle +sinXsinYsinZcosXcosYcosZ+sinYsin^2Zcos^2XcosY\;\; ?$
$\displaystyle 6sinXsinYsinZcosXcosYcosZ$
$\displaystyle \le\ sinXcosX\left(sin^2Zcos^2Y+sin^2Ycos^2Z\right)$
$\displaystyle +sinYcosY\left(sin^2Xcos^2Z+sin^2Zcos^2X\right)$
$\displaystyle +sinZcosZ\left(sin^2Xcos^2Y+Sin^2Ycos^2X\right)$
This inequality holds if the following 3 inequalities hold
$\displaystyle (1)\;\;\;2sinYsinZcosYcosZ\ \le\ (sinZcosY)^2+(sinYcosZ)^2$
$\displaystyle (2)\;\;\;2sinXsinZcosXcosZ\ \le\ (sinXcosZ)^2+(sinZcosX)^2$
$\displaystyle (3)\;\;\;2sinXsinYcosXcosY\ \le\ (sinXcosY)^2+(sinYcosX)^2$
$\displaystyle (1)\;\;\;(sinZcosY-sinYcosZ)^2\ \ge\ 0\;\;?\;\;\;Yes$
$\displaystyle (2)\;\;\;(sinXcosZ-sinZcosX)^2\ \ge\ 0\;\;?\;\;\;Yes$
$\displaystyle (3)\;\;\;(sinXcosY-sinYcosX)^2\ \ge\ 0\;\;?\;\;\;Yes$
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 $\displaystyle \sin^2 \frac{B+C}2\geqslant \sin B\sin C$. If we let $\displaystyle \beta = B/2$ and $\displaystyle \gamma = C/2$ then that inequality becomes $\displaystyle (\sin\beta\cos\gamma+\cos\beta\sin\gamma)^2 \geqslant 4\sin\beta\cos\beta\sin\gamma\cos\gamma$. But that is equivalent to $\displaystyle (\sin\beta\cos\gamma-\cos\beta\sin\gamma)^2 \geqslant 0$, which is obviously true.
When we had
$\displaystyle 8sinXsinYsinZcosXcosYcosZ\ \le\ 2sinXsinYsinZcosXcosYcosZ+........\;\;?$
we were left with (due to the common terms)
$\displaystyle 6sinXsinYsinZcosXcosYcosZ$
$\displaystyle \le\ sinXcosX\left(sin^2Zcos^2Y+sin^2Ycos^2Z\right)$
$\displaystyle +sinYcosY\left(sin^2Xcos^2Z+sin^2Zcos^2X\right)$
$\displaystyle +sinZcosZ\left(sin^2Xcos^2Y+sin^2Ycos^2X\right)\;\ ;?$
Now we split $\displaystyle 6sinXsinYsinZcosXcosYcosZ$ into 3 of $\displaystyle 2sinXsinYsinZcosXcosYcosZ$
and compare each of these to the components on the right-hand side of the inequality.
For the first of the 3 inequalities
$\displaystyle sinXcosX(2sinYsinZcosYcosZ)\;\le\ sinXcosX\left(sin^2Zcos^2Y+sin^2Ycos^2Z\right)\;\; ?$
Dropping the common factor gives
$\displaystyle 2sinYsinZcosYcosZ\ \le\ sin^2Zcos^2Y+sin^2Ycos^2Z\;\;?$
and similarly for the other two.