Cách 1: Áp dụng BĐT Cauchy

\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=\dfrac{2}{b}\)

Tương tự: \(\dfrac{b}{c^2}+\dfrac{1}{b}\ge\dfrac{2}{c}\)

\(\dfrac{c}{a^2}+\dfrac{1}{c}\ge\dfrac{2}{a}\)

Cộng vế theo vế các BĐT vừa chứng minh rồi rút gọn, ta có:

\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)(đpcm)

Đẳng thức xảy ra khi \(a=b=c\)

Cách 2: Áp dụng BĐT Bunyakovsky

\(\left(\dfrac{\sqrt{a}}{b}.\dfrac{1}{\sqrt{a}}+\dfrac{\sqrt{b}}{c}.\dfrac{1}{\sqrt{b}}+\dfrac{\sqrt{c}}{a}.\dfrac{1}{\sqrt{c}}\right)^2\le\left(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\ge\left(\dfrac{a}{b^2}+\dfrac{b}{a^2}+\dfrac{c}{a^2}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\)(đpcm)

Đẳng thức xảy ra khi \(a=b=c\)