I can't solve the following problem:
Show that amongst all triangles with perimeter $3p,$ the equilateral triangle with side $p$ has the largest area. Further show that $9p^2\ge 12\sqrt{3}\Delta.$
I have got that the area is maximum when $ab+bc+ca$ is maximum and $abc$ is minimum (by using Heron's Formula). I am stuck at this point.
Any help will be appreciated.
By Heron's formula $$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{1}$$ and by the AM-GM inequality $$ (-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3 \tag{2} $$ Equality is attained only at $a=b=c$.