Bich Ngoc 2k1

$\sum \frac{1}{a+c+2(b+c)}< \sum \frac{1}{4}.\left ( \frac{1}{a+c}+\frac{1}{2(b+c)} \right )=\frac{3}{4}.\sum \frac{1}{a+b}< \frac{3}{16}.\left ( \sum \frac{1}{a} \right )=\frac{3}{16}$

chỗ $\frac{3}{4} là \frac{3}{8}$ mới đúng chứ