Problem :
Prove that :
All triangle have same perimeter the largest space is Equilateral triangles
I know that :
area $S=\sqrt{s(s-a)(s-b)(s-c)}$
With : $s=\frac{a+b+c}{2}$
Also I'm going to use this inequality
$(a+b+c)^{3}≥9abc$ , but I don't know how I use it ??
I have already to see your hints and ideas ?
Hint: Use $$\sqrt[4]{\frac 13s(s-a)(s-b)(s-c)}\leq \frac {\frac 13s+(s-a)+(s-b)+(s-c)}4,$$ where equality holds when $a=b=c$.
The area of the triangle is
$$S= \frac12 r (a+b+c)=\frac12 rp\tag 1$$
Also,
$$p=2r\left(\cot \frac A2 + \cot \frac B2 + \cot \frac C2\right)\tag 2$$
Combine (1) and (2) to get
$$S= \frac{p^2}4\cdot \frac 1{\cot \frac A2 + \cot \frac B2 + \cot \frac C2}$$
Since $\cot \theta $ is a convex function over $\theta\in (0,\frac\pi2)$, the following holds according to Jensen's inequality,
$$\cot \frac A2 + \cot \frac B2 + \cot \frac C2 \ge 3\cot\frac{A+B+C}6=3\cot\frac{180^\circ}6=3\sqrt3$$
where the equality holds for $A=B=C = 60^\circ$. Thus, the largest area, given by $S_m=\frac{p^2}{12\sqrt3}$, comes from the equilateral triangle.
The area is given by $\sqrt{s(s-a)(s-b)(s-c)}$ where $(a,b,c)$ is in the triangle with vertices $(2s,0,0), (0,2s,0), (0,0,2s).$ Since a continuous real-valued function on a compact set attains its maximum, we see that the problem has a solution.
Suppose the triangle with maximum area has sides $a,b,c$. If we keep side $a$ fixed, and allow the other two sides to vary, their sum must be $2s-a$. Thus, if we place two of the vertices at $P_1=(-\frac{a}{2},0)$ and $P_2=(\frac{a}{2},0)$ the third vertex $(x,y)$ will lie on an ellipse with foci $P_1, P_2$, and the area of the triangle will be $\frac{a}{2}|y|$ and will be greatest when $x=0$, so the triangle is isosceles.
Since whatever side of the triangle we choose, the other two sides are equal, the triangle is equilateral.
