Áp dụng bđt AM - GM ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=2\left|\dfrac{a}{c}\right|\ge2\dfrac{a}{c}\)(1)
\(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{c^2}{a^2}}=2\left|\dfrac{c}{b}\right|\ge2\dfrac{c}{b}\)(2)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{b^2}{c^2}.\dfrac{c^2}{a^2}}=2\left|\dfrac{b}{a}\right|\ge2\dfrac{b}{a}\)(3)
Cộng vế với vế của (1);(2);(3) ta được :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\)(đpcm)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
