Sửa đề: \(1< \dfrac{a}{a+b+c}+\dfrac{b}{a+b+d}+\dfrac{c}{a+c+d}+\dfrac{d}{b+c+d}< 2\)
Ta có : \(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\) (1)
\(\dfrac{b}{a+b+d}>\dfrac{b}{a+b+c+d}\) (2)
\(\dfrac{c}{a+c+d}>\dfrac{c}{a+b+c+d}\) (3)
\(\dfrac{d}{c+b+d}>\dfrac{d}{a+b+c+d}\) (4)
Từ (1)(2)(3)(4) =>\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+d}+\dfrac{c}{a+c+d}+\dfrac{d}{b+c+d}>\dfrac{a+b+c+d}{a+b+c+d}=1\)
Lại có:\(\dfrac{a}{a+b+c}< \dfrac{a+d}{a+b+c+d}\)(Vì a<a+b+c)
\(\dfrac{b}{a+b+d}< \dfrac{b+c}{a+b+c+d}\)(Vì b<a+b+d)
\(\dfrac{c}{a+c+d}< \dfrac{b+c}{a+b+c+d}\)(Vì c<c+a+d)
\(\dfrac{d}{b+c+d}< \dfrac{d+a}{a+b+c+d}\)(Vì d<d+b+c)
=>\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+d}+\dfrac{c}{a+c+d}+\dfrac{d}{b+c+d}< \dfrac{2\left(a+b+c+d\right)}{a+b+c+d}=2\\ \text{Vậy 1< ...< 2}\)
