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I can't solve the following problem:

Show that amongst all triangles with perimeter $3p,$ the equilateral triangle with side $p$ has the largest area. Further show that $9p^2\ge 12\sqrt{3}\Delta.$

I have got that the area is maximum when $ab+bc+ca$ is maximum and $abc$ is minimum (by using Heron's Formula). I am stuck at this point.

Any help will be appreciated.

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  • $\begingroup$ What is $\Delta$? $\endgroup$ – Roman83 May 23 '16 at 14:33
  • $\begingroup$ $\Delta$ is area of the triangle. $\endgroup$ – user333900 May 23 '16 at 14:33
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By Heron's formula $$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{1}$$ and by the AM-GM inequality $$ (-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3 \tag{2} $$ Equality is attained only at $a=b=c$.

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  • $\begingroup$ (+1) A good and elegant solution, Thanks. But the second part of the question? $\endgroup$ – user333900 May 23 '16 at 14:43
  • $\begingroup$ @user333900: for the second part you just have to put $a=b=c=p$, getting $16\Delta^2 \leq 3p^4$. $\endgroup$ – Jack D'Aurizio May 23 '16 at 14:44
  • $\begingroup$ @Jack D'Aurizio It isn't said that $a=b=c=p$. $p$ is arbitrary. $\endgroup$ – user333900 May 23 '16 at 14:47
  • $\begingroup$ @user333900: you have that the perimeter is fixed, equal to $3p$, and equality in $(2)$ is attained only at $a=b=c$. Given $a+b+c=3p$, it follows that $a=b=c=p$. $\endgroup$ – Jack D'Aurizio May 23 '16 at 14:49
  • $\begingroup$ @Jack D'Aurizio No, I meant for all $p$ such that $3p$ is perimeter of triangle with area $\Delta$ prove that $9p^2\ge 12\sqrt{3}\Delta$. $\endgroup$ – user333900 May 23 '16 at 15:17

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