Question

CM BĐT : \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}.\)

Answer 1

Áp dụng BĐT : \(x^2+y^2\ge2xy\) ( xảy ra đẳng thức khi \(x=y\) ) , ta có :

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2.\frac{a}{b}.\frac{b}{c}=2.\frac{a}{c}.\) Tương tự : \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2.\frac{b}{a};\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2.\frac{c}{b}\)

Cộng từng vế ba bất đẳng thức trên :

\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\)

\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{c}+\frac{b}{a}+\frac{c}{b}.\)