Câu hỏi của Hoàng Đỗ Việt

Question

Cho số nguyên dương a, b, c, dChứng tỏ rằng: $1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2$

ST · Accepted Answer

Ta có: $\frac{a}{a+b+c}>\frac{a}{a+b+c+d}$ $\frac{b}{b+c+d}>\frac{b}{a+d+c+d}$ $\frac{c}{c+d+a}>\frac{c}{a+b+c+d}$ $\frac{d}{d+a+b}>\frac{d}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+b+a}+\frac{d}{d+a+b}< \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}>\frac{a+b+c+d}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 1$ (1)Lại có: $\frac{a}{a+b+c}< \frac{a+c}{a+b+c+d}$ $\frac{b}{b+c+d}< \frac{b+d}{a+b+c+d}$ $\frac{c}{c+d+a}< \frac{c+a}{a+b+c+d}$ $\frac{d}{d+a+b}< \frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{a+c}{a+b+c+d}+\frac{b+d}{a+b+c+d}+\frac{c+a}{a+b+c+d}+\frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{2a+2b+2c+2d}{a+b+c+d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2$ (2)Từ (1)(2) => $1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2$ (đpcm) Câu trả lời: Ta có: $\frac{a}{a+b+c}>\frac{a}{a+b+c+d}$ $\frac{b}{b+c+d}>\frac{b}{a+d+c+d}$ $\frac{c}{c+d+a}>\frac{c}{a+b+c+d}$ $\frac{d}{d+a+b}>\frac{d}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+b+a}+\frac{d}{d+a+b}< \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}>\frac{a+b+c+d}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 1$ (1)Lại có: $\frac{a}{a+b+c}< \frac{a+c}{a+b+c+d}$ $\frac{b}{b+c+d}< \frac{b+d}{a+b+c+d}$ $\frac{c}{c+d+a}< \frac{c+a}{a+b+c+d}$ $\frac{d}{d+a+b}< \frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{a+c}{a+b+c+d}+\frac{b+d}{a+b+c+d}+\frac{c+a}{a+b+c+d}+\frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< \frac{2a+2b+2c+2d}{a+b+c+d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2$$\Rightarrow\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2$ (2)Từ (1)(2) => $1< \frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}< 2$ (đpcm)

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