Simple proof that equilateral triangles have maximum area

You can use Heron's formula for the area of a triangle to help here. $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ where $s$ is the semi-perimeter $$s = \frac{a+b+c}{2}$$ To find the conditions for the maximum area, we want to compute derivatives w.r.t. $a$, $b$ and $c$, but since they are not independent variables, we first substitute the semi-perimeter equation into the are equation, to get rid of $c$. $$c = 2s-a-b$$ $$\therefore A = \sqrt{s(s-a)(s-b)(a+b-s)}$$ Setting the derivative w.r.t. $a$ to zero: $$\frac{\text{d}A}{\text{d}a} = \frac{s(s-b)(2s-2a-b)}{2A} = 0$$ $$\Rightarrow 2s-2a-b = 0$$ (Note that $b=s$ is not a solution since $A \ne 0 \Rightarrow b \ne s$.)

Similarly, setting the derivative w.r.t. $b$ to zero yields $$2s-2b-a = 0$$ Solving simultaneously gives $$a = b = \frac{2s}{3}$$ and substituting back gives $$c = \frac{2s}{3}$$ So $a=b=c$ and the triangle is equilateral.

As Carl pointed out, Heron's formula gives the area of a triangle with sides $a$, $b$, and $c$ as $A = \sqrt{s(s-a)(s-b)(s-c)}$, where $s=\displaystyle\frac{1}{2}(a+b+c)$.

Let $2s=a+b+c>0$, and suppose $A^2=s(s−a)(s−b)(s−c)$ is maximized for $a$, $b$, and $c$ that are not all equal. Without loss of generality, assume $a<b$.

It's straightforward to show that $s(s−\displaystyle\frac{a+b}{2})(s−\frac{a+b}{2})(s−c)>A^2$, which means that you'll get a triangle of larger area if you replace the unequal sides $a$ and $b$ with equal sides, keeping the same perimeter. This contradicts the assumption that $a\neq b$ at the maximum.

For what it's worth, here's another proof without calculus.

Fix a perimeter $p$, and suppose that $a$, $b$, and $c=p-a-b$ are the side lengths of a largest triangle with perimeter $p$. Let $F_1$, $F_2$, and $P$, respectively, be the vertices of the triangle opposite the sides of length $a$, $b$, and $c$.

Given $c$, we'll show that it must be the case that $a=b$, proving that the largest triangle is isosceles. Repeating the argument given $a$ or $b$ will prove it must be equilateral.

Place $F_1$ and $F_2$ on the $x$-axis, and consider where $P$ can lie. Wherever it is, the sum of the distances from $P$ to the two points $F_1$ and $F_2$ is $a+b$, which is constant, so the locus of possible locations of $P$ is an ellipse with foci $F_1$ and $F_2$. The area of $F_1F_2C$ is $c|y|$, where $y$ is the $y$-coordinate of $P$, so the triangle's area is greatest when $P$ is at a point on the ellipse as far from the $x$-axis as possible. Therefore $P$ is at an endpoint of the ellipse's vertical axis, and by symmetry, $a=b$.

Here's a picture with $P$ not at the optimal position.