Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\)
\(\Rightarrow xyz=\frac{a}{b}.\frac{b}{c}.\frac{c}{a}=1\)
Bất đẳng thức đã cho tương đương với: \(\Leftrightarrow x^2+y^2+z^2\ge\frac{z}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow2.\left(x^2+y^2+z^2\right)-2.\left(xy+yz+zx\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\left(\forall x;y;z\right)\)
Dấu "=" xảy ra khi \(\Leftrightarrow x=y=z\Rightarrow a=b=c\left(đpcm\right)\)