## Task $3 / 89$

Let $A$ be the area and $u=a+b+c$ the perimeter of a triangle with side lengths $a, b$, and $c$. Determine the maximum of the ratio $z=\frac{A}{u^{2}}$.

For which triangles is it assumed?
According to Heron's formula for the area of a triangle, $A^{2}=s(s-a)(s-b)(s-c)$, where $s=\frac{u}{2}$, thus $u=2 s$. According to the inequality between the geometric mean and the arithmetic mean,

$$
\sqrt[3]{\frac{A^{2}}{s}}=\sqrt[3]{(s-a)(s-b)(s-c)} \leq \frac{s-a+s-b+s-c}{3}=\frac{s}{3}
$$

From this, it follows that

$$
\frac{A^{2}}{s} \leq \frac{s^{3}}{27}, \quad A^{2} \leq \frac{s^{4}}{27}=\frac{u^{4}}{16 \cdot 27}, \quad \frac{A}{u^{2}} \leq \frac{1}{12 \sqrt{3}}=\frac{\sqrt{3}}{36}
$$

Since the equality in the inequality holds if and only if $a=b=c$, the maximum is achieved for the equilateral triangle.