Ta có :
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\\\dfrac{b}{b+c}>\dfrac{b}{a+b+c}\\\dfrac{c}{c+a}>\dfrac{c}{a+b+c}\end{matrix}\right.\) \(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a+b+c}{a+b+c}=1\left(1\right)\)
Ta lại có :
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\\\dfrac{b}{b+c}< \dfrac{b+a}{a+b+c}\\\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\end{matrix}\right.\) \(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)
mà \(1;2\) là 2 số nguyên liên tiếp
\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\notin Z\left(đpcm\right)\)
