Ta có : \(\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\)
\(\dfrac{b}{b+c}>\dfrac{b}{b+a+c}\)
\(\dfrac{c}{c+a}>\dfrac{c}{c+a+b}\)
=> M > \(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)(1)
Lại có : \(\dfrac{a}{a+b}< \dfrac{a+b}{a+b+c}\)
\(\dfrac{b}{b+c}< \dfrac{b+c}{b+c+a}\)
\(\dfrac{c}{c+a}< \dfrac{c+a}{a+b+c}\)
=> M < \(\dfrac{a+b+b+c+c+a}{a+b+c}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)(2)
Từ (1) và (2) suy ra : 1<M<2
=> M không là số nguyên
Vậy................