Find Minimum area of given hexagon. Geometry Question. It has been a year Since I am searching for an answer to this question.
This question was probably asked in International Mathematics Olympiad but I am not a 100% sure.
Q : ABCDEF Is a concave hexagon. Let P be a random point on inside hexagon. Suppose $[APB] = 4$ , $[CPD] = 6$ and $[EPF] = 9$. (Here $[ABC]$ denotes the area of triangle $ABC$). What is the minimum possible area of hexagon $ABCDEF$ ?

 A: At least one among triangles $APB$, $CPD$ or $EPF$ must lie entirely inside the hexagon, so its area is at least $4$. On the other hand, the difference between the hexagon area and $4$ can be made as small as one wishes, as shown for example in the picture below.

In this case we have a degenerate hexagon whose area is $4+\epsilon$, and $\epsilon$ can be taken arbitrarily small. So the infimum of the area is 4.
EDIT
As Peter Shor's comment points out, my assumption that one among triangles $APB$, $CPD$ or $EPF$ must lie entirely inside the hexagon was wrong: his construction shows that the area of the hexagon can be vanishingly small. 
In the picture below I followed his idea, but with a non-degenerate hexagon. Start the construction with an equilateral triangle $A'E'C'$ of side $2\sqrt3\epsilon$, so that its center $P$ is at a distance $\epsilon$ from its sides. 
Draw then points $A$, $C$ and $E$ on the sides of $A'E'C'$ such that
$AA'=CC'=EE'=\epsilon^2$. Finally, draw $B$, $D$ and $F$ on the extensions of the sides of $A'E'C'$ such that $AB=8/\epsilon$, $CD=12/\epsilon$ and $EF=18/\epsilon$. 
The conditions on the areas of $PAB$, $PCD$ and $PEF$ are satisfied, while
the area of hexagon $ABCDEF$ turns out to be 
$$
area={19\over3}\sqrt3\epsilon+3\sqrt3\epsilon^2
-{9\over2}\epsilon^3+{3\over4}\sqrt3\epsilon^4.
$$

This tends to zero as $\epsilon\to0$, so the infimum of hexagon area is zero.
