cách khác:$\frac{a}{2b+2c-a}+\frac{b}{2a+2c-b}=\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2ab+2bc-b^2}\geq \frac{(a+b)^2}{2.2ab-(a^2+b^2)+2ac+2bc}\geq \frac{(a+b)^2}{2(a^2+b^2)-(a^2+b^2)+2ac+2bc}=\frac{(a+b)^2}{a^2+b^2+2ac+2bc}$
tương tự:....
$\sum_{cyc}^{}\frac{a}{2b+2c-a}\geq \frac{2(a+b+c)^2}{2(a^2+b^2+c^2)+4(ab+bc+ac)}=\frac{(a+b+c)^2}{(a+b+c)^2}=1 $