$A= ( 1+\frac{1}{a}).( 1+\frac{1}{b}).( 1+\frac{1}{c})$
$= ( 1+\frac{a+b+c}{a}).( 1+\frac{a+b+c}{b}).( 1+\frac{a+b+c}{c})$
$= ( 2+\frac{b}{a}+\frac{c}{a}).( 2+\frac{a}{b}+\frac{c}{b}).( 2+\frac{b}{c}+\frac{a}{c})$
Áp dụng bđt cô si, ta có:
$A≥ ( 2+2.\frac{\sqrt[]{bc}}{a}).( 2+2.\frac{\sqrt[]{ac}}{b}).( 2+2.\frac{\sqrt[]{ab}}{c})$
$≥ 4.\sqrt[]{\frac{\sqrt[]{bc}}{a}}.4.\sqrt[]{\frac{\sqrt[]{ac}}{b}}.4.\sqrt[]{\frac{\sqrt[]{ab}}{c}}$
$⇔ A≥ 64 (đpcm)$
Dấu = xảy ra khi a=b=c=$\frac{1}{3}$
