
A3. Find the maximum value of the area of a triangle having side lengths $a, b, c$ with

$$
a^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}
$$


## Solution

Without any loss of generality, we may assume that $a \leq b \leq c$.

On the one hand, Tchebyshev's inequality gives

$$
(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \leq 3\left(a^{3}+b^{3}+c^{3}\right)
$$

Therefore using the given equation we get

$$
a+b+c \leq 3 \text { or } p \leq \frac{3}{2}
$$

where $p$ denotes the semi perimeter of the triangle.

On the other hand,

$$
p=(p-a)+(p-b)+(p-c) \geq 3 \sqrt[3]{(p-a)(p-b)(p-c)}
$$

Hence

$$
\begin{aligned}
p^{3} \geq 27(p-a)(p-b)(p-c) & \Leftrightarrow p^{4} \geq 27 p(p-a)(p-b)(p-c) \\
& \Leftrightarrow p^{2} \geq 3 \sqrt{3} \cdot S
\end{aligned}
$$

where $S$ is the area of the triangle.

Thus $S \leq \frac{\sqrt{3}}{4}$ and equality holds whenever when $a=b=c=1$.

## Comment

Cauchy's inequality implies the following two inequalities are true:

$$
\frac{a+b+c}{3} \leq \frac{a^{2}+b^{2}+c^{2}}{a+b+c} \leq \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}
$$

Now note that

$$
\frac{a+b+c}{3} \leq \frac{a^{2}+b^{2}+c^{2}}{a+b+c}
$$

gives

$$
(a+b+c)^{2} \leq 3\left(a^{2}+b^{2}+c^{2}\right)
$$

whereas $\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \leq \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}$, because of our assumptions, becomes $\frac{a^{2}+b^{2}+c^{2}}{a+b+c} \leq 1$, and so,

$$
a^{2}+b^{2}+c^{2} \leq a+b+c
$$

Combining (1) and (2) we get

$(a+b+c)^{2} \leq 3(a+b+c)$ and then $a+b+c \leq 3$.
