Theo bất đẳng thức $AM-GM$ ta được:
$\begin{array}{l} \left\{ \begin{array}{l} \dfrac{{{a^2}}}{b} + b \ge 2\sqrt {\dfrac{{{a^2}}}{b}.b} = 2a\left( 1 \right)\\ \dfrac{{{b^2}}}{c} + c \ge 2\sqrt {\dfrac{{{b^2}}}{c}.c} = 2b\left( 2 \right)\\ \dfrac{{{c^2}}}{a} + a \ge 2\sqrt {\dfrac{{{c^2}}}{a}.a} = 2a\left( 3 \right) \end{array} \right.\\ \left( 1 \right) + \left( 2 \right) + \left( 3 \right)\\ \Rightarrow \dfrac{{{a^2}}}{b} + \dfrac{{{b^2}}}{c} + \dfrac{{{c^2}}}{a} + \left( {a + b + c} \right) \ge 2\left( {a + b + c} \right)\\ \Leftrightarrow \dfrac{{{a^2}}}{b} + \dfrac{{{b^2}}}{c} + \dfrac{{{c^2}}}{a} \ge a + b + c \end{array}$
Dấu bằng xảy ra khi và chỉ khi $a=b=c$
