Áp dụng bđt Cauchy-schwarz dạng engel và bđt \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\) ta có:
\(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}=\frac{a^2b^2}{abc}+\frac{b^2c^2}{abc}+\frac{c^2a^2}{abc}\ge\frac{\left(ab+bc+ca\right)^2}{3abc}\ge\frac{3\left(ab^2c+abc^2+a^2bc\right)}{3abc}=a+b+c\)
Dấu "=" \(\Leftrightarrow a=b=c\)