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\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)
\(\Rightarrow ad+ab< bc+ab\)
\(\Rightarrow a.\left(b+d\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}\left(1\right)\)
\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)
\(\Rightarrow ad+cd< bc+cd\)
\(\Rightarrow d.\left(a+c\right)< c.\left(b+d\right)\)
\(\Rightarrow\frac{d}{c}< \frac{b+d}{a+c}\)
\(\Rightarrow\frac{c}{d}>\frac{a+c}{b+d}\left(2\right)\)
Từ (1) và (2) ,suy ra đpcm
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) \(\Rightarrowđpcm\)
\(\frac{a}{b}< \frac{c}{d}\)\(\Rightarrow ad< bc\)\(\Rightarrow ad+ab< bc+ab\)\(\Rightarrow a.\left(b+d\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\)
\(\frac{a}{b}< \frac{c}{d}\)\(\Rightarrow ad< bc\)\(\Rightarrow ad+cd< bc+cd\)\(\Rightarrow d.\left(a+c\right)< c.\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\)
Có \(\frac{a}{b}< \frac{c}{d}\left(b,d>0\right)\)
\(\Rightarrow ad< bc\)
\(\Rightarrow ab+ad< ab+bc\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (vì b, b + d > 0) (1)
Có \(ad< bc\)
\(\Rightarrow ad+cd< bc+cd\)
\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\) (vì b + d, d > 0) (2)
Từ (1)(2) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
\(\text{Ta có:}\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{a}{b}.\frac{d}{c}< \frac{c}{d}.\frac{d}{c}\)
\(\Rightarrow\frac{ad}{bc}< \frac{cd}{dc}\)
\(\Rightarrow\frac{ad}{bc}< 1\)
\(\Rightarrow ad< 1.bc\)
\(\Rightarrow ad< bc\)
\(\cdot\text{Từ }ad< bc\Rightarrow ad+ab< bc+ab\)
\(\Rightarrow a.\left(b+d\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)
\(\cdot\text{Từ }ad< bc\Rightarrow ad+cd< bc+cd\)
\(\Rightarrow d.\left(a+c\right)< c.\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)
\(\text{Từ (1) và (2)}\Rightarrow\frac{a}{c}< \frac{a+c}{b+d}< \frac{c}{d}\left(đpcm\right)\)