đặt $1-a=x;1-b=y;1-c=z$ $\Rightarrow x,y,z>0$ và $x+y+z=2$
S=$(1+\frac{2}{x})(1+\frac{2}{y})(1+\frac{2}{z})=1+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{4}{xy}+\frac{4}{yz}+\frac{4}{zx}+\frac{8}{xyz}$
$ \geq 1+\frac{18}{x+y+z}+\frac{16}{xyz}\geq 1+9+\frac{16}{(\frac{x+y+z}{3})^{3}}=64$
dấu '=" $\Leftrightarrow a=b=c=\frac{1}{3}$