Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$

Applying the inequality $AM \ge GM $ on {$ab,bc,ca$}, we get:
$$\frac{ab+bc+ca}{3} \ge (a^2.b^2.c^2)^{\frac13}$$
$\Rightarrow$ $(abc)^{\frac23} \le \frac{12}{3} = 4$
i.e. the maximum value of abc is $4^{\frac32}$ = $8$.
Also, maximum value of $8$ is attained for $a=b=c=2$
