cho a,b,c >0 chứng minh rằng
a/(b+c) + b/(c+a) + c/(a+b) >3/2
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} = \frac{a^{2}}{ab+bc}+\frac{b^{2}}{bc+ab}+\frac{c^{2}}{ca+bc} \geq \frac{(a+b+c)^{2}}{2(ab+bc+ca)} \geq \frac{(a+b+c)^{2}}{2.\frac{1}{3}(a+b+c)^{2}} = \frac{3}{2}$