Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9 
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$

This is how I tried doing it:
Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.
Using AM>GM for $p, q$, I get:
$$\frac{p+q}{2} > {(pq)}^{1/2}$$
$$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$
And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$
Thus, 
 $$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)<9, $$
which is the opposite of what had to be proven. What did I do wrong?

By the AM-GM inequality,
$$\frac{a}{b}+\frac{b}{a}\geq 2 $$
and so on, so:
$$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 3+\sum_{cyc}\left(\frac{b}{a}+\frac{a}{b}\right) \geq 3+6 = \color{red}{9}$$
as wanted.
