Ta thấy: \(\frac{a}{a+b}>\frac{a}{a+b+c}\)
\(\frac{b}{b+c}>\frac{b}{a+b+c}\)
\(\frac{c}{c+a}>\frac{c}{a+b+c}\)
=>\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)
=>\(M>1\) (1)
Lại có:
Áp dụng: Với \(\frac{m}{n}<1=>\frac{m}{n}<\frac{m+k}{n+k}\)(với \(m,n\in N\cdot\))
Ta có: \(\frac{a}{a+b}<1=>\frac{a}{a+b}<\frac{a+c}{a+b+c}\)
\(\frac{b}{b+c}<1=>\frac{b}{b+c}<\frac{b+a}{a+b+c}\)
\(\frac{c}{c+a}<1=>\frac{c}{c+a}<\frac{c+b}{a+b+c}\)
=>\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}\)
=>\(M<\frac{2.\left(a+b+c\right)}{a+b+c}\)
=>M<2 (2)
Từ (1) và (2)
=>1<M<2
=>M không phải số nguyên
=>ĐPCM
