Denote  \(C=\dfrac{a}{a+b+c}+\dfrac{b}{b+c+d}+\dfrac{c}{c+d+a}+\dfrac{d}{d+a+b}\)

+) We have :

\(\dfrac{a}{a+b+c}>\dfrac{a}{a+b+c+d}\)

\(\dfrac{b}{b+c+d}>\dfrac{b}{a+b+c+d}\)

\(\dfrac{c}{c+d+a}>\dfrac{c}{a+b+c+d}\)

\(\dfrac{d}{d+a+b}>\dfrac{d}{a+b+c+d}\)

So \(C>\dfrac{a}{a+b+c+d}+\dfrac{b}{a+b+c+d}+\dfrac{c}{a+b+c+d}+\dfrac{d}{a+b+c+d}=\dfrac{a+b+c+d}{a+b+c+d}=1\)

+) We have this too :v 

\(\dfrac{a}{a+b+c}< \dfrac{a}{a+c}\)

\(\dfrac{b}{b+c+d}< \dfrac{b}{b+d}\)

\(\dfrac{c}{c+d+a}< \dfrac{c}{c+a}\)

\(\dfrac{d}{d+a+b}< \dfrac{d}{b+d}\)

Plus equality to equality :

\(\Rightarrow C< \dfrac{a}{a+c}+\dfrac{b}{b+d}+\dfrac{c}{a+c}+\dfrac{d}{b+d}=\dfrac{a+c}{a+c}+\dfrac{b+d}{b+d}=2\) 

Finally , \(1< C< 2\) :) 

Denote  C=aa+b+c+bb+c+d+cc+d+a+dd+a+b

+) We have :

aa+b+c>aa+b+c+d

bb+c+d>ba+b+c+d

cc+d+a>ca+b+c+d

dd+a+b>da+b+c+d

So C>aa+b+c+d+ba+b+c+d+ca+b+c+d+da+b+c+d=a+b+c+da+b+c+d=1

+) We have this too :v 

aa+b+c<aa+c

bb+c+d<bb+d

cc+d+a<cc+a

dd+a+b<db+d

Plus equality to equality :

⇒C<aa+c+bb+d+ca+c+db+d=a+ca+c+b+db+d=2

Finally , 1<C<2

 :)