Question

Cho các số thực dương \(a,b,c\) thỏa mãn \(a+b+c=3\) . Chứng minh rằng 

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{6abc}{ab+bc+ca}\ge5\)

Answer 1

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>=3\sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}}=3\)

\(ab+ac+bc< =3\sqrt{ab\cdot ac\cdot bc}=3abc\)

=>6abc/ab+ac+bc>=2

=>a/b+b/c+c/a+6ab/ab+ac+bc>=5

Answer 2

\(ab+ac+bc\le3\sqrt{ab\cdot ac\cdot bc}\) là sao?