$A \ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c}-\frac{2}{abc}$Do $\frac{1}{3}\left( \frac 1a+\frac 1b+\frac 1c \right) \ge \frac{3}{a+b+c}$
Nên chỉ cần cm $\frac{2}{3}\left( \frac 1a+\frac 1b+\frac 1c \right) \ge \frac{2}{abc}$
$\Leftrightarrow ab+bc+ca \ge 3$
Bdt cuối đúng do $(ab+bc+ca)^2 \ge 3abc(a+b+c) \ge3(ab+bc+ca)$