đặt 1-a=x;1-b=y;1-c=z \Rightarrow x,y,z>0 và x+y+z=2S=(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{2.9}{x+y+z}+\frac{16}{xyz}\geq 1+9+\frac{16}{(\frac{x+y+z}{3})}{3})^{3}}=64 dấu '=" \Leftrightarrow a=b=c=\frac{1}{3}
đặt
1-a=x;1-b=y;1-c=z \Rightarrow x,y,z>0 và
x+y+z=2S=
(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} $\geq
1+\frac{2.9}{x+y+z}+\frac{16}{xyz}
\geq 1+9+\frac{16}{(\frac{x+y+z}{3})}{3})^{3}}=64
dấu '=" \Leftrightarrow a=b=c=\frac{1}{3}$