I'll give the same answer I did on Stackoverflow.
I believe you can solve this problem in linear time, except that you will have to order your set of triangle length first, which is $\Theta(n\cdot \log n)$.
Here's how I would do it.
// V is the set of length of triangle.
// Heron Formula
define A(a,b,c) = SQRT(4a^2*b^2 - (a^2 + b^2 - c^2) ^2) / 4
// I is the last position in V
I := V.length - 1
// T(I) is the area of Maximum Area Triangle
if V[I]^2 <= V[I-1]^2+V[I-2]ˆ2
then return A(V[I-2],V[I-1],V[I])
return MAX( A(V[I-2],V[I-1],V[I]), T(I-1) )
Explanation: Assuming that $a \le b \le c$, when $c^2\le a^2+b^2$, if you increase any of the variable (a, b or c), you will have a triangle that has a larger area. So, if you peak the three largest numbers and the equation $c^2\le a^2+b^2$ is true, that means you have found the largest area. But, if $c^2\le a^2+b^2$ is not true, that is $c^2>a^2+b^2$, then a triangle with a different setup might have a greater area. If we change $b$ or $a$ with a lower value, we will have lower area, so that is not an option. So, the only option we have is to choose a different value for $c$, which is $T(I-1)$, which is also a recursion. The maximum value between $A(V[I-2],V[I-1],V[I])$ and $T(I-1)$ will be your answer.