TITLE: Simple proof that equilateral triangles have maximum area
QUESTION [14 upvotes]: Using Lagrange multipliers, it can be shown that a triangle with given perimeter has the maximum possible area, if it is equilateral. Is there a simple geometric proof of that fact ?

REPLY [6 votes]: assume that a,b,c are the sides of a triangle, we know that: 
$$P=\sqrt{S\left(S-a\right)\left(S-b\right)\left(S-c\right)} ~~\mbox{  (Heron's formula)}$$
and:
$$2S=a+b+c$$
$$S=\frac{a+b+c}{2}$$
Due to the inequality of arithmetic and geometric we can say:
$$\frac{(S-a)+(S-b)+(S-c)}{3}\ge\sqrt[3]{(S-a)(S-b)(S-c)}$$
$$=\frac{3S-(a+b+c)}{3}\ge\sqrt[3]{(S-a)(S-b)(S-c)}$$
$$=\frac{3S-2S}{3}\ge\sqrt[3]{(S-a)(S-b)(S-c)}$$
$$=\frac{S}{3}\ge\sqrt[3]{(S-a)(S-b)(S-c)}$$
$$=\frac{S^3}{27}\ge{(S-a)(S-b)(S-c)}$$
$$=\frac{S^4}{27}\ge{S(S-a)(S-b)(S-c)}$$
$$=\frac{S^2}{3\sqrt{3}}\ge\sqrt{S\left(S-a\right)\left(S-b\right)\left(S-c\right)}$$
now we proved that the maximum area in this case cannot exceed : $$\frac{S^2}{3\sqrt{3}}$$
and for the equality situation $(S-a)=(S-b)=(S-c)$ so $a=b=c$ and our triangle is equilateral .