\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\) (do a,b,c >0)
Ta có đpcm
#)Giải :
Ta có :
\(\hept{\begin{cases}\frac{a}{a+b}>\frac{a}{a+b+c}\\\frac{b}{b+c}>\frac{b}{a+b+c}\\\frac{c}{c+a}>\frac{c}{a+b+c}\end{cases}}\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1\left(đpcm\right)\)
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1\left(đk:a,b,c>0\right)\)
ta có \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{b}{b+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}>\frac{a+b+c}{a+b+c}=1\left(1\right)\)
từ (1) =>............
sửa chút nha
ta có \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)chứ ko pk\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{b}{b+c}\)
nha
