\(\dfrac{1,28}{3,15}=\dfrac{128}{315}\) ; \(\dfrac{\dfrac{2}{5}}{3\dfrac{1}{4}}=\dfrac{8}{65}\) ; \(\dfrac{-1\dfrac{3}{7}}{1,24}=\dfrac{-250}{217}\)

Xét : \(\dfrac{a^2}{b}+b=\dfrac{a^2+b^2}{b}\ge\dfrac{2ab}{b}=2a\)

^{\(\Rightarrow\dfrac{a^2}{b}+b\ge2a\)}

^{Tương tự ta có : \(\dfrac{b^2}{c}+c\ge2a;\dfrac{c^2}{a}+a\ge2c\)}

Cộng theo vế cac BPT trên :

\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)