Question

Cho a,b,c>0.CMR :

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

Answer 1

Áp dụng BĐT AM-GM ta có: 

\(\frac{a}{b^2}+\frac{1}{a}\ge2\sqrt{\frac{a}{b^2}\cdot\frac{1}{a}}=2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)

\(\frac{b}{c^2}+\frac{1}{b}\ge2\sqrt{\frac{b}{c^2}\cdot\frac{1}{b}}=\frac{2}{c}\)

\(\frac{c}{a^2}+\frac{1}{c}\ge2\sqrt{\frac{c}{a^2}\cdot\frac{1}{c}}=\frac{2}{a}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\Leftrightarrow VT\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Dấu "=" xảy ra khi \(a=b=c\)

Answer 2

Đáp án của tôi giống Thắng Nguyễn