Minimizing Perimeter Of A Triangle

The minimum perimeter for a given area will be obtained with an equilateral triangle. Similar ideas tell us that the minimum perimeter of a rectangle of a given area is a square, and the minimum surface area of a spheroid of a given volume is a sphere. That is why soap bubbles form spheres instead of any other possible shape. The area of an equilateral triangle is given by $a =$ $\frac {\sqrt {3}} {4}s$ . The side here is 1 and the perimeter 3.

Let $p$ , $q$ and $r$ denote the sides of the trinagle, 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]

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

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

$\implies s^2_{min} = \dfrac{\sqrt{3}}{4} \times 3\sqrt{3}$ [This minimum occurs when $p=q=r$ , or the triangle is equilateral]

$\implies s^2_{min} = \dfrac{9}{4}$

$\implies s_{min} = \dfrac{3}{2}$

$\implies 2s_{min} = \boxed{3}$ .