Question

chứng minh bđt:

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

Answer 1

Áp dụng BĐT Cô - si : x² + y² ≥ 2xy

=> \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\) ≥ \(2.\dfrac{a}{c}\) ( 1)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\) ≥ \(2.\dfrac{b}{a}\) ( 2)

\(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}\) ≥ \(2.\dfrac{c}{b}\) ( 3)

Cộng từng vế của ( 1 , 3 , 3) , ta có :

\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\) ≥ \(2.\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)

=> ĐPCM