Consider nine numbers $\frac{a}{2},\frac{a}{2},\frac{b}{3},\frac{b}{3},\frac{b}{3},\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$

Using $\mathrm{A}\mathrm{M}\ge \mathrm{G}\mathrm{M}$, we get,
$\frac{\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}\mathrm{}}{9}\ge {\left({\left(\frac{a}{2}\right)}^2{\left(\frac{b}{3}\right)}^3{\left(\frac{c}{4}\right)}^4\right)}^{\frac{1}{9}}$
$\Rightarrow \frac{a+b+c}{9}\ge {\left(\frac{a^2{b}^3{c}^4}{2^2{3}^3{4}^4}\right)}^{\frac{1}{9}}$
$\Rightarrow {\left(\frac{27}{9}\right)}^9\ge \frac{a^2{b}^3{c}^4}{2^2{3}^3{2}^8}$
$\Rightarrow a^2{b}^3{c}^4\le 2^{10}\cdot 3^{12}$