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?
 A: You are using $x+1/x\ge 2$ in the opposite direction. You can do it by AM-GM as follows:
$$a+b+c>3\sqrt[3]{abc}\ \ \land\ \ 1/a+1/b+1/c>3/\sqrt[3]{abc}\ \ \implies\ \ (a+b+c)(1/a+1/b+1/c)>9$$
Alternatively, you can use the Cauchy-Schwarz inequality:
$$(a+b+c)(1/a+1/b+1/c)>(1+1+1)^2=9$$
A: 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.
A: Well, although everyone has answered it, but I think $A.M \geq H.M.$ should also do the work. It's more better in this case. Moreover no one has addressed as to why the OP is wrong.
We know that $\frac{a+b+c}{3} \geq \frac{3}{\frac1a + \frac1b + \frac1c}$.
Henceforth, we get $(a+b+c)(\frac1a + \frac1b + \frac1c) \geq 9$.
Coming to the error in your attempt, in the step where you have written $$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$
Clearly, $x+\frac1x \geq 2$, which gives the RHS of the above inequality greater than 3. So therefore, you have said that if $A>B$ and $A>C$ then $B>C$, which is of course not true. Here we cannot compare $B$ and $C$. Hence we couldn't proceed by your method.
A: Consider
$$(a-b)^2 \ge 0$$
$$a^2-2ab+b^2 \ge 0$$
$$a^2+b^2 \ge 2ab$$
$$\frac{a^2+b^2}{ab} \ge 2$$
$$\frac{a}{b}+\frac{b}{a}\ge 2$$
without a loss of generality, we can also say that
$$\frac{a}{c}+\frac{c}{a}\ge 2$$
and
$$\frac{b}{c}+\frac{c}{b}\ge 2$$
of course
$$3=1+1+1=\frac{a}{a}+\frac{b}{b}+\frac{c}{c}$$
So, putting it all together we have
$$\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\ge 3+2+2+2$$
$$(a+b+c)\bigg(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\ge 9$$
