$\text{Ta có hằng đẳng thức }$
$\text{(a+b+c)³=a³+b³+c³+3.(a+b).(b+c).(c+a)}$
$\text{Áp dụng AM-GM ,ta có}$
$\text{a+b≥2.$\sqrt[]{ab}$ }$
$\text{b+c≥2.$\sqrt[]{cb}$ }$
$\text{c+a≥2.$\sqrt[]{ac}$ }$
$\text{⇒(a+b).(b+c).(c+a)≥8abc}$
$\text{⇔3.(a+b).(b+c).(c+a)≥24abc}$
$\text{Suy ra }$
$\text{(a+b+c)³≥a³+b³+c³+24abc }$