Chứng minh M=a/(a+b)+b/(b+c)+c/(a+c) không là số nguyên biết a, b, c khác không

\(M=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\\\dfrac{b}{b+c}>\dfrac{b}{a+b+c}\\\dfrac{c}{c+a}>\dfrac{c}{a+b+c}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}\)

Mà \(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow M>1\left(1\right)\)

Lại có: \(\left\{{}\begin{matrix}\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\\\dfrac{b}{b+c}< \dfrac{b+a}{a+b+c}\\\dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+a}{a+b+c}\)

Mà: \(\dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{a+c+b+a+c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow M< 2\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)

\(\Rightarrow1< M< 2\)

Vậy \(M=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\) không phải là số nguyên (Đpcm)