$= \frac{{{a^3} - {a^2}b - a{b^2} + {b^3}}}{b} = \frac{{{{\left( {a - b} \right)}^2}.\left( {a + b} \right)}}{b} \ge 0$(vì $a,b > 0)$

Vậy $\frac{{{a^3}}}{b} \ge {a^2} + ab - {b^2}$

Chứng minh tương tự : $\frac{{{b^3}}}{c} \ge {b^2} + bc - {c^2} & & & \frac{{{c^3}}}{b} \ge {c^2} + ca - {a^2}$

$\Rightarrow \frac{{{a^3}}}{b} + \frac{{{b^3}}}{c} + \frac{{{c^3}}}{a} \ge ab + bc + ca$

Dấu xảy ra $\Leftrightarrow a = b = c$