Cx ko hẳn là xưa lắm!!
Ko mất tính tq, g/s rằng $a\geq b\geq c$, khi đó:
$\frac{a(b+c)}{a^2+bc}\geq \frac{a(b+c)}{a^2+ac}=\frac{b+c}{a+c}\geq \frac{b}{a}$
và:
$\frac{c(a+b)}{c^2+ab}\geq \frac{c}{b}$
Vậy:
$\Sigma \frac{a(b+c)}{a^2+bc}\geq \frac{b}{a}+\frac{ab}{b^2+ca}+\frac{c}{b}=\frac{b^2+ca}{ab}+\frac{ab}{b^2+ca}\geq 2$
$\rightarrow $.......