If A is the area and 2s is the sum of the sides of a triangle, then

$$\begin{gathered} \mathrm{We} \mathrm{have} 2\mathrm{s}=\mathrm{a}+\mathrm{b}+\mathrm{c}, {\mathrm{A}}^2=\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right) \\ \boxed{A.M\ge G.M} \\ \Rightarrow \frac{\left(s-a\right)+\left(s-b\right)+\left(s-c\right)}{3}\ge {\left(\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)\right)}^{1/3} \\ \Rightarrow \frac{\mathrm{s}}{3}\ge {\left[\frac{{\mathrm{A}}^2}{\mathrm{s}}\right]}^{\frac{1}{3}}\Rightarrow \frac{s^3}{27}⩾\frac{A^2}{s} \\ \Rightarrow A^2\le \frac{s^4}{27} \\ \therefore \mathrm{A}\le \frac{{\mathrm{s}}^2}{3\sqrt{3}} \end{gathered}$$