Since $1 \geq \sqrt[3]{abc} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$ we have \[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3 \] We also have \[\frac{6}{a+b+c} \leq \frac{2}{3} \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\] So we are done since \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 1 + \frac{2}{3} \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\]

By AM-GM, $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{3}{\sqrt[3]{abc}}\ge 1+\frac{2}{\sqrt[3]{abc}}\ge 1+\frac{6}{a+b+c}.$

1. Let $a,\,b,\,c>0$. Prove that $\left( a+b+c \right)+\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\ge \left( 1+\frac{6}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \right)+\left( 1+\frac{6}{a+b+c} \right)$. 2. Let $a,\,b,\,c>0\,,\,abc=1$. Prove that $\left( a+b+c \right)\cdot \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)\ge \left( 1+\frac{6}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \right)\cdot \left( 1+\frac{6}{a+b+c} \right)$. 3. Find the smallest real positive number $k$ such that $\left( a+b+c+k \right)\cdot \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+k \right)\ge \left( 1+\frac{6}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \right)\cdot \left( 1+\frac{6}{a+b+c} \right),\,\forall \,a,b,c>0$

Now, $1\ge\sqrt[3]{abc}\overset{\text{GM-HM}}{\ge}\frac{3}{\sum_{cyc}\frac{1}{a}}\Longrightarrow \sum_{cyc}\frac{1}{a}\ge3$ furthermore $2/3\sum_{cyc}\frac{1}{a}\overset{\text{AM-HM}}{\ge}\frac{6}{\sum_{cyc}a}$, and finally combining the last two inequalities we obtain the desired result $\blacksquare$.