Đáp án:
\[\dfrac{1}{a}+\dfrac 1b+\dfrac 1c\ge 9\qquad(*)\] Vì $a+b+c=1\to (*)\Leftrightarrow \dfrac 1a +\dfrac 1b + \dfrac 1c\ge \dfrac{9}{a+b+c}(**)$
Áp dụng AM-GM:
\[a+b+c\geq 3\sqrt[3]{abc}\] \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt[3]{\frac{1}{abc}}\]
\[\to \dfrac 1a +\dfrac 1b + \dfrac 1c\ge \dfrac{9}{a+b+c}\]
\[\to \dfrac 1a+\dfrac 1b+\dfrac 1c\ge 9\]
