Câu hỏi của dangminhhieu

Question

Với a, b, c, d >0 CMR    1<$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2$

Reina · Accepted Answer

Ta có: $\frac{a}{b+c+d}>\frac{a}{a+b+c+d};\frac{b}{a+c+d}>\frac{b}{a+c+b+d};\frac{c}{b+c+d}>\frac{c}{a+b+c+d}$$\frac{d}{a+b+c}>\frac{d}{a+b+c+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+c+b+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}>\frac{a+b+c+d}{a+b+c+d}=1\left(1\right)$Vì $\frac{a}{b+c+d}< 1\Rightarrow\frac{a}{b+c+d}< \frac{a+c}{b+c+a+d}$$\frac{b}{c+d+a}< 1\Rightarrow\frac{b}{b+c}< \frac{b+a}{a+b+c+d}$$\frac{c}{b+c+d}< 1\Rightarrow\frac{c}{b+c+d}< \frac{c+b}{a+b+c+d}$$\frac{d}{a+b+c}< 1\Rightarrow\frac{d}{a+b+c}< \frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< \frac{a+c}{a+b+c+d}+\frac{b+a}{a+b+c+d}+\frac{c+d}{a+b+c+d}+\frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< \frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\left(2\right)$$\left(1\right)\left(2\right)\Rightarrow1< \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2$Vậy a,b,c,d>0 thì $1< \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2\left(đpcm\right)$Câu trả lời: Ta có: $\frac{a}{b+c+d}>\frac{a}{a+b+c+d};\frac{b}{a+c+d}>\frac{b}{a+c+b+d};\frac{c}{b+c+d}>\frac{c}{a+b+c+d}$$\frac{d}{a+b+c}>\frac{d}{a+b+c+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+c+b+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}>\frac{a+b+c+d}{a+b+c+d}=1\left(1\right)$Vì $\frac{a}{b+c+d}< 1\Rightarrow\frac{a}{b+c+d}< \frac{a+c}{b+c+a+d}$$\frac{b}{c+d+a}< 1\Rightarrow\frac{b}{b+c}< \frac{b+a}{a+b+c+d}$$\frac{c}{b+c+d}< 1\Rightarrow\frac{c}{b+c+d}< \frac{c+b}{a+b+c+d}$$\frac{d}{a+b+c}< 1\Rightarrow\frac{d}{a+b+c}< \frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< \frac{a+c}{a+b+c+d}+\frac{b+a}{a+b+c+d}+\frac{c+d}{a+b+c+d}+\frac{d+b}{a+b+c+d}$$\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< \frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\left(2\right)$$\left(1\right)\left(2\right)\Rightarrow1< \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2$Vậy a,b,c,d>0 thì $1< \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2\left(đpcm\right)$

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP