+) Ta có : $\dfrac{a}{a+b+c} > \dfrac{a}{a+b+c+d} $
(Vì :$ a,b,c,d>0$ )
$\dfrac{b}{b+c+d}>\dfrac{b}{a+b+c+d}$
$\dfrac{c}{c+d+a}>\dfrac{c}{a+b+c+d}$
$\dfrac{d}{d+a+b}>\dfrac{d}{a+b+c+d}$
Nên $\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}>\dfrac{a+b+c+d}{a+b+c+d} =1$ (1)
+) Lại có : $\dfrac{a}{a+b+c} <\dfrac{a+d}{a+b+c+d} $
Tương tự ta có :
$\dfrac{b}{b+c+d}<\dfrac{b+a}{a+b+c+d}$
$\dfrac{c}{c+d+a}>\dfrac{c+b}{a+b+c+d}$
$\dfrac{d}{d+a+b}>\dfrac{d+c}{a+b+c+d}$
Nên $\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}<\dfrac{2.(a+b+c+d)}{a+b+c+d} =2$ (2)
+) Từ $(1),(2)$
$⇒1<\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}<2$
