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15 tháng 3 2017

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)

            

DD
31 tháng 5 2021

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

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

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

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

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

Suy ra đpcm. 

15 tháng 2 2020

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

Ta có: \(\frac{a}{a+b+c}< \frac{a}{a+c}\)

\(\frac{b}{b+c+d}< \frac{b}{b+d}\)

\(\frac{c}{c+d+a}< \frac{c}{a+c}\)

\(\frac{d}{d+a+b}< \frac{d}{d+b}\)

\(\Rightarrow S< \left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+d}+\frac{d}{d+b}\right)\)

\(\Rightarrow S< 2\left(1\right)\)

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

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

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

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

\(\Rightarrow S>1\left(2\right)\)

Từ (1) và (2) \(\Rightarrowđpcm\)

15 tháng 2 2020

nhanh the