Question

Giúp e với

Answer 1

 

\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge2\\ \Rightarrow\dfrac{1}{a+1}\ge\left(1-\dfrac{1}{b+1}\right)+\left(1-\dfrac{1}{c+1}\right)\\ \Rightarrow\dfrac{1}{a+1}\ge\dfrac{b}{b+1}+\dfrac{c}{c+1}\)

Theo BĐT AM-GM ; ta có :

\(\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge 2\sqrt{\dfrac{bc}{\left(b+1\right)\left(c+1\right)}}\\ \Rightarrow\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{8abc}{\left(a+1\right)\left(b+c\right)\left(c+1\right)}\\ \Rightarrow a.b.c\le\dfrac{1}{8}\)