Ta có:
\(\left(\dfrac{a}{b}+\dfrac{b}{c}\right)^2\ge0\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+2.\dfrac{a}{b}.\dfrac{b}{c}\ge0\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge\dfrac{2a}{c}\left(1\right)\)
Tương tự:
\(\left(\dfrac{b}{c}+\dfrac{c}{a}\right)^2\ge0\Rightarrow\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2b}{a}\left(2\right)\)
\(\left(\dfrac{a}{b}+\dfrac{c}{a}\right)^2\ge0\Rightarrow\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\ge\dfrac{2c}{b}\left(3\right)\)
Từ (1)(2)(3) cộng vế theo vế ta được:
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\)
