\(m=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
=>\(\dfrac{a}{a+b}>\dfrac{a}{a+b+c};\dfrac{b}{b+c}>\dfrac{b}{a+b+c};\dfrac{c}{c+a}>\dfrac{c}{a+b+c}\)
=>\(M=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a+b+c}{a+b+c}\)
ta có \(\dfrac{a}{a+b}< \dfrac{a+c}{a+b+c};\dfrac{b}{b+c}< \dfrac{a+b}{a+b+c};\dfrac{c}{c+a}< \dfrac{b+c}{a+b+c}\)
cộng biểu thức ta đc \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}+\dfrac{a+b}{a+b+c}\)
=>\(M< \dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
=>1<M<2
=> M KHÔNG LÀ SỐ NGUYÊN
\(M=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
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)
