Question: Prove that an equilateral triangle allows for the greatest packing density when only packing one circle into a triangle. 

I have though of starting with the unit circle inscribed in an equilateral triangle, then using the ratio of the area of the circle to the left over area of the triangle to construct the proof. I have seen the length of the sides of the triangle as 2v3, but I am not exactly sure what that means. I am not sure what to do from here, any tips on how to determine the area of the triangle? 
Would there be any other ways of constructing this proof differently?

Any advise or suggestions would be appreciated!

Answer: 
The area of a triangle $\Delta = rs$, where $r$ is its inradius and $s$ is its semiperimeter. The area of the incircle is $\pi r^2$. We want to maximize the ratio of the circle's area to the triangle's area; namely, the ratio $$\begin{align}\frac{\pi r^2}{rs} &= \frac{\pi r}{s} \\[4pt] &\propto \frac{r}{s} \end{align}$$

From $rs = \Delta = \sqrt{s(s-a)(s-b)(s-c)}$ where $a$, $b$, $c$ are the sides of the triangle, we obtain $r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$ whereupon
$$\frac{r}{s} = \sqrt{\frac{(s-a)(s-b)(s-c)}{s^3}}$$

Since the quantity under the sqaure root is positive, maximizing $\frac{r}{s}$ is equivalent to maximizing $\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}$. By the AM-GM inequality,
$$\frac{(s-a) + (s-b) + (s-c)}{3} \geq \sqrt[3]{(s-a)(s-b)(s-c)} \\
\implies \frac{(s-a) + (s-b) + (s-c)}{3s} \geq \sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}$$
but the left-hand side is merely $\frac{1}{3}$, so
$$\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}} \leq \frac{1}{3}$$

Note that equality occurs only when $s - a = s - b = s - c \implies a = b = c$, or in other words, for an equilateral triangle. Thus, only when the triangle is equilateral, the maximum value of $\sqrt[3]{\frac{(s-a)(s-b)(s-c)}{s^3}}$ and thereby of $\frac{r}{s}$ and $\frac{\pi r^2}{rs}$ is obtained.

(*Extra:* the actual value of $\frac{\pi r^2}{rs}$ is slightly greater than $0.6$ for an equilateral triangle. The incircle covers slightly more than $60\%$ of the triangle's area.)
Let's consider the dual problem: Find the triangle $\triangle$ of minimal area circumscribing the unit circle with center $O$. Let $C\_i$ $(1\leq i\leq3)$ be the points of contact. If $\angle(C\_{i-1}OC\_i)=2\phi\_i<\pi$ then area of the triangle is given by
$${\rm area}(\triangle)=\tan\phi\_1+\tan\phi\_2+\tan\phi\_3\ .$$
Now the function $\phi\mapsto\tan\phi$ is convex when $0<\phi<{\pi\over2}$, and this implies
$${\tan\phi\_1+\tan\phi\_2+\tan\phi\_3\over3}\geq\tan{\phi\_1+\phi\_2+\phi\_3\over3}=\tan{\pi\over3}=\sqrt{3}\ .$$
It follows that
$${\rm area}(\triangle)\geq3\sqrt{3}$$
with equality sign for the equilateral triangle. The unit disk then consumes ${\pi\over3\sqrt{3}}=60.46\%$ of the triangle area.