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 $a=b=c=1$.