# Find the minimum value of the perimeter

Find the minimum value of the perimeter of a triangle whose area is 3 $$cm^2$$ I tried it using Hero rule $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ $$9 = s(s-a)(s-b)(s-c)$$ But it did not serve me well ?

• Indeed; so, try to prove that the minimum is obtained when the triangle is equilateral. Geometric reasoning may serve you better than Heron's formula. – Gerry Myerson Jun 26 at 2:41

By AM-GM and by your work we obtain: $$9=s(s-a)(s-b)(s-c)\leq s\left(\frac{s-a+s-b+s-c}{3}\right)^3=\frac{s^4}{27}.$$ Can you end it now?
• How did you get the last equality to $\frac{s^4}{27}$? – Don Draper Jun 26 at 3:05
• @Don Draper Because $s-a+s-b+s-c=3s-(a+b+c)=3s-2s=s.$ – Michael Rozenberg Jun 26 at 3:07
• Sorry, at first I didn't notice that $s$ doesn't stand for the triangle area. – Don Draper Jun 26 at 3:20