\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\left(\cdot\right)\)
Áp dụng BĐT Nê bít, ta có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\le\dfrac{3}{2}< 2\left(\cdot\cdot\right)\)
Từ \(\left(\cdot\right)\left(\cdot\cdot\right)\)
\(\Rightarrow1< \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}< 2\left(đpcm\right)\)

