Question

Cho a, b, c >0 thỏa mãn :

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

CM : M không có giá trị nguyên.

Answer 1

Ta có:

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)\(\Rightarrow\)\(M>1\left(1\right)\)

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

= \(3-\left(\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{c+a}\right)< 2\) \(\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{c+a}>1\)

(Vì \(\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{c+a}>1\)

\(\Rightarrow1< M< 2\)

Vậy M không có giá trị nguyên(đpcm)