Question

Chứng minh rằng với mọi số a,b,c thuộc Z* ta có:

\(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

(giải chi tiết vào vs nhá - thanks trước)

Answer 1

Đặt \(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

Vì \(a,b,c\in Z\)*

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

\(\Rightarrow A>\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow A>1\)

Vì \(a,b,c\in Z\)*

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

\(\Rightarrow A< \dfrac{a+c+b+a+c+b}{a+b+c}=2\)

\(\Rightarrow A< 2\)

Vậy \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

Answer 2

Đặt \(A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

Vì \(a,b,c\in Z\)*

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

Nên: \(\dfrac{b}{b+c}>\dfrac{b}{b+c+a};\dfrac{c}{c+a}>\dfrac{c}{c+a+b}\)

Vậy\(A=\)\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow A>1\)(1)

Lại có :\(\dfrac{a}{a+b}< \dfrac{a+b}{a+b+c}\)

Nên:\(\dfrac{b}{b+c}< \dfrac{b+c}{b+c+a};\dfrac{c}{c+a}< \dfrac{c+a}{c+a+b}\)

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

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

\(\Rightarrow A< 2\)(2)

Từ (1) và (2) \(\Rightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

\(\Rightarrow A\) không phải là số nguyên