ta có
\(\dfrac{a}{a+b}>\dfrac{a}{a+b+c}\)
\(\dfrac{b}{b+c}>\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+c}>\dfrac{c}{a+b+c}\)
cộng vế với vế ta đc
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}>\dfrac{a+b+c}{a+b+c}\)
<=> \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}>1\left(đpcm\right)\)