Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$ 
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$.

Should I use AM-GM for the expression in the middle of the inequality? We have $a+b \geq 2\sqrt{ab}$ etc.?

To prove:
$2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) \geq \frac{9}{a+b+c}$,
let $x=a+b, y=b+c, z=c+a$, then we have:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Rightarrow 3+(\frac{y}{x}+\frac{x}{y}) +(\frac{y}{z}+\frac{z}{y})+ (\frac{z}{x}+\frac{x}{z}) \geq 3+2+2+2=9$
