Đáp án:
Có : \(\left\{ \begin{array}{l}\dfrac{a}{a + b + c } < \dfrac{a}{a+b }\\ \dfrac{b}{a + b + c} < \dfrac{b}{b + c}\\ \dfrac{c}{a+b+c}<\dfrac{c}{c+a}\end{array} \right.\)
Cộng theo vế ta được :
`⇔ a/(a + b+c) + b/(a + b + c) + c/(a + b + c) < a/(a + b) + b/(b + c) + c/(c + a)`
`⇔ (a + b + c)/(a + b + c) < a/(a + b) + b/(b + c) + c/(c + a)`
`⇔ 1 < a/(a + b) + b/(b + c) + c/(c + a)` `(1)`
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Mặt khác có : \(\left\{ \begin{array}{l}\dfrac{a}{a+b}<\dfrac{a + c}{a + b +c}\\ \dfrac{b}{b + c}<\dfrac{a + b}{a + b + c}\\ \dfrac{c}{a+c}<\dfrac{b+c}{a+b+c}\end{array} \right.\)
Cộng theo vế ta được :
`⇔ a/(a + b) + b/(b + c) + c/(a + c) < (a + c)/(a + b + c) + (a + b)/(a + b + c) + (b + c)/(a + b + c)`
`⇔ a/(a + b) + b/(b + c) + c/(a + c) < (a + c + a +b + b + c)/(a + b + c)`
`⇔ a/(a + b) + b/(b + c) + c/(a + c) < (2a + 2c + 2b)/(a + b + c)`
`⇔ a/(a + b) + b/(b + c) + c/(a + c) < (2 (a + b+c) )/(a + b +c)`
`⇔ a/(a + b) + b/(b + c) + c/(a + c) < 2` `(2)`
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Từ `(1), (2)`
`->1 < a/(a + b) + b/(b + c) + c/(c + a) < 2`
