Question

Cho a, b, c > 0 và \(a+b+c\le3\) . CMR :

A= \(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{3}{2}\)

Answer 1

Với a ; b ; c > 0 ; AD BĐT Cauchy ta được : 

\(A=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{3+ab+bc+ac}\ge\)   

\(\ge\dfrac{9}{3+\dfrac{\left(a+b+c\right)^2}{3}}\ge\dfrac{9}{3+\dfrac{3^2}{3}}=\dfrac{3}{2}\)

" = " \(\Leftrightarrow a=b=c=1\)