\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)
=>M>1 (1)
\(M=\left(1-\frac{b}{a+b}\right)+\left(1-\frac{c}{b+c}\right)+\left(1-\frac{a}{c+a}\right)<3-\left(\frac{b}{a+b+c}+\frac{c}{b+c+a}+\frac{c}{c+a+b}\right)\)
\(\Rightarrow M<3-1=2\Rightarrow M<2\) (2)
từ (1);(2)=>1<M<2
=>M ko là 1 số nguyên(đpcm) ,với a,b,c>0