Question

Cho M = \(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+d}+\dfrac{c}{b+c+d}+\dfrac{d}{a+d+c}\) (a,b,c thuộc N sao)

CMR: M ko là số tự nhiên

Answer 1

Lời giải:

Ta có:

\(M=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{a+d+c}\)

\(> \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)

\(\Leftrightarrow M>\frac{a+b+c+d}{a+b+c+d}=1(1)\)

Mặt khác:

\(M=1-\frac{b+c}{a+b+c}+1-\frac{a+d}{a+b+d}+1-\frac{b+d}{b+c+d}+1-\frac{a+c}{a+d+c}\)

\(\Leftrightarrow M=4-\underbrace{\left(\frac{b+c}{a+b+c}+\frac{a+d}{a+b+d}+\frac{b+d}{b+c+d}+\frac{a+c}{a+d+c}\right)}_{N}\)

Có: \(N>\frac{b+c}{a+b+c+d}+\frac{a+d}{a+b+c+d}+\frac{b+d}{a+b+c+d}+\frac{a+c}{a+b+c+d}\)

\(\Leftrightarrow N>\frac{2(a+b+c+d)}{a+b+c+d}=2\)

\(\Rightarrow M=4-N< 4-2\Leftrightarrow M< 2(2)\)

Từ \((1);(2)\Rightarrow 1< M< 2\Rightarrow M\not\in \mathbb{N}\)