Áp dụng bđt AM-GM: \(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2a}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\) \(\dfrac{b}{c^2}+\dfrac{1}{b}\ge2\sqrt{\dfrac{b}{c^2b}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\) \(\dfrac{c}{a^2}+\dfrac{1}{c}\ge2\sqrt{\dfrac{c}{a^2c}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\) Cộng theo vế: \(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\Leftrightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)Dấu "=" xảy ra khi: \(a=b=c\)