Với các số thực dương $a;\,b;\,c$, chứng minh
- $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge 1+\dfrac{a^2+b^2+c^2}{2(ab+bc+ca)}.$
- $\dfrac{a}{c+a}+\dfrac{b}{a+b}+\dfrac{c}{b+c}\ge 1+\dfrac{ab+bc+ca}{a^2+b^2+c^2+ab+bc+ca}.$
- $\dfrac{a}{c+a}+\dfrac{b}{a+b}+\dfrac{c}{b+c}\ge \dfrac{a+b+c}{a+b+c-\sqrt[3]{abc}}.$
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]