2009 Zhautykov International Olympiad Problems/Problem 3
Problem
For a convex hexagon with an area
, prove that:

Lemma
First, let's set aside the original problem and introduce the "Gergonne point" and its distance to the vertices.
Let the incircle of touch
,
,
at
,
,
respectively. By Ceva's theorem, it is easy to see that
,
,
are concurrent at a point
, which we call the Gergonne point of
.
Let ,
,
. As shown in the figure, by Stewart's theorem, we have
In
, by Menelaus' theorem, we have
Solving this, we get
Noting that by AM-GM inequality, we have
Thus, we obtain the following lemma.
Lemma: Let be the Gergonne point of
, then
Returning to the original problem
Let be the Gergonne point of
. As shown in the figure, by the lemma, we have
Similarly, we have
,
, adding these together, we get the desired result.