Maximum Area with A Given Perimeter

Let $p$ , $q$ and $r$ denote the sides of $\Delta ABC$ , and $2s=(p+q+r)$ denote the perimeter, $\Delta$ denote the area.

By AM-GM inequality, we have,

$\dfrac{(s-p)+(s-q)+(s-r)}{3} \geq \sqrt[3]{(s-p)(s-q)(s-r)}$ [With equality holds only when $s-p=s-q=s-r$ , or $p = q = r$ ]

$\Leftrightarrow \dfrac{s}{3} \geq \sqrt[3]{(s-p)(s-q)(s-r)}$

$\Leftrightarrow \dfrac{s^3}{27} \geq (s-p)(s-q)(s-r)$

$\Leftrightarrow \dfrac{s^4}{27} \geq s(s-p)(s-q)(s-r)$

$\Leftrightarrow \dfrac{s^2}{3\sqrt{3}} \geq \sqrt{ s(s-p)(s-q)(s-r)}$ [Taking square root on both sides]

$\Leftrightarrow \dfrac{s^2}{3\sqrt{3}} \geq \Delta$

$\implies \Delta_{max} = \dfrac{s^2}{3\sqrt{3}}$ [This maximum occurs when $p=q=r$ , or the triangle is equilateral]

$\implies \Delta_{max} = \dfrac{(2s)^2}{4\times 3\sqrt{3}} = \dfrac{(2s)^2}{12\sqrt{3}} = \dfrac{2017^2}{12\sqrt{3}}$ .

Now, $2017^2$ and $12$ are coprime to each other; $3$ is square-free. So, $a=2017^2$ , $b=12$ and $c=3$ .

Therefore, $(a+b+c)=2017^2+12+3=\boxed{4068304}.$