Có $F=\frac{36a}{bc}+\frac{2b}{ac}+\frac{c}{ab}=(\frac{9a}{bc}+\frac{b}{ac})+(\frac{b}{ac}+\frac{c}{ab})+\frac{27a}{bc}\geq \frac{6}{c}+\frac{2}{a}+\frac{27a}{bc}$$=(\frac{2}{a}+\frac{18a}{bc})+\frac{6}{c}+\frac{9a}{bc}\geq \frac{12}{\sqrt{bc}}+\frac{6}{c}+\frac{9a}{bc}\geq \frac{12}{\sqrt{3.3}}+\frac{6}{3}+\frac{9.1}{3.3}=7$
Dấu = có khi a=1;b=c=3