Heron's formula can be rearranged into 16A^2 = 2a^2b^2 + 2a^2c^2 + 2b^2c^2 - a^4 - b^4 - c^4. Finding the maximum value of 16A^2 will also maximize A.
The condition 2a^2 + bc = 6 implies a^2 = 3 - bc/2. Substitute this into the area equation to yield 16A^2 = -b^4 - b^3c + 7b^2c^2/4 - bc^3 - c^4 + 6b^2 + 3bc + 6c^2 - 9.
The symmetry of b and c implies that the maximum will occur when b=c. Then 16A^2 = -9b^4/4 + 15b^2 - 9.
The first derivative of the right side is -9b^3+30b, which has one positive root b=sqrt(10/3), which implies c=sqrt(10/3), a=sqrt(4/3), and A=sqrt(146/144).
Possible solution
