Question

Cho các số thực dương a,b,c thỏa a+b+c=1

CMR: \(\dfrac{a}{1+bc}+\dfrac{b}{1+ca}+\dfrac{c}{1+ab}\ge\dfrac{9}{10}\)

help me pls!!

Answer 1

\(VT=\dfrac{a^2}{a+abc}+\dfrac{b^2}{b+abc}+\dfrac{c^2}{c+abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3abc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\dfrac{1}{9}\left(a+b+c\right)^3}=\dfrac{1^2}{1+\dfrac{1}{9}.1^3}=\dfrac{9}{10}\)

Answer 2

=9/10