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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 ?

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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jun 26 at 2:41
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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?

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  • $\begingroup$ How did you get the last equality to $\frac{s^4}{27}$? $\endgroup$ – Don Draper Jun 26 at 3:05
  • $\begingroup$ @Don Draper Because $s-a+s-b+s-c=3s-(a+b+c)=3s-2s=s.$ $\endgroup$ – Michael Rozenberg Jun 26 at 3:07
  • $\begingroup$ Sorry, at first I didn't notice that $s$ doesn't stand for the triangle area. $\endgroup$ – Don Draper Jun 26 at 3:20

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