given area of a triangle, minimum value of its perimeter stack exchange

Question

given area of traingle $\sqrt3 $ , find minimum value of it's perimeter.

I didn't get any idea where to start,

Answer 1

We know with Heron's formula (let $s=\frac{a+b+c}{2}$, half of the perimeter):

$$\sqrt3=\sqrt{s(s-a)(s-b)(s-c)}$$

And so

$$3=s(s-a)(s-b)(s-c)\tag{$*$}$$

With AM-GM, we see

\begin{align} \frac13s&=\frac{(s-a)+(s-b)+(s-c)}3\\ &\geq \sqrt[3]{(s-a)(s-b)(s-c)} \end{align}

And so

$$\frac{1}{27}s^3\geq(s-a)(s-b)(s-c)$$

which becomes, when multiplying both sides by $s$:

$$\frac{1}{27}s^4\geq s(s-a)(s-b)(s-c)=3$$

simplify this to see

$$s^4\geq 81=3^4$$

Which we can take the fourth root of to get $s\geq 3$. Equality in AM-GM holds iff all the terms are equal, so $s-a=s-b=s-c$ which happens iff $a=b=c$, thus, $a=b=c=2$ (to satisfy equation $(*)$). The minimum perimiter is $6$ (since $s\geq3$, and $s$ is half the perimeter) and being reached by an equilateral triangle with side lengths $2$.