Đặt \(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
Ta có:
\(A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}\)\(=\dfrac{a+b+c}{a+b+c}=1\left(1\right)\)
\(A< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}\)
\(=\dfrac{a+b+b+c+c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\)
Từ \((1);(2)\) ta có \(1< A< 2\)
Vậy \(A\) không phải là số nguyên