Giải thích các bước giải:
Đặt $A=\dfrac a{a+b}+\dfrac b{b+c}+\dfrac c{c+a}$
Ta có :
$a<a+b\to \dfrac{a}{a+b+c}<\dfrac{a}{a+b}<\dfrac{a+c}{a+b+c}$
Tương tự
$\dfrac{b}{a+b+c}<\dfrac{b}{b+c}<\dfrac{b+a}{a+b+c}$
$\dfrac{c}{a+b+c}<\dfrac{c}{c+a}<\dfrac{c+b}{a+b+c}$
$\to \dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}<A<\dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}$
$\to 1<A<2$
