Chào em, em tham khảo:
Giải thích các bước giải:
Ta thấy:
`+)` $\dfrac{a}{a+b+c}<\dfrac{a}{b+c}<\dfrac{a+a}{a+b+c}$
`+)` $\dfrac{b}{a+b+c}<\dfrac{b}{a+c}<\dfrac{b+b}{a+b+c}$
`+)` $\dfrac{c}{a+b+c}<\dfrac{c}{a+b}<\dfrac{c+c}{a+b+c}$
Cộng vế với vế ta được:
$\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}<\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}<\dfrac{a+a}{a+b+c}+\dfrac{b+b}{a+b+c}+\dfrac{c+c}{a+b+c}$
$⇒\dfrac{a+b+c}{a+b+c}<\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}<\dfrac{2a+2b+2c}{a+b+c}$
$⇒\dfrac{a+b+c}{a+b+c}<\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}<\dfrac{2(a+b+c)}{a+b+c}$
$⇒1<\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}<2$
Vậy $1<\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}<2$.
