Áp dụng BĐT Cô-si:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{a^2}{b^2}.\dfrac{b^2}{c^2}}=\dfrac{2a}{c}\)
Tương tự: \(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\ge\dfrac{2c}{b}\) ; \(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{2b}{a}\)
Cộng vế:
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\right)\)
\(\Rightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{c}\)
Dấu "=" xảy ra khi \(a=b=c\)
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