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.?
 A: Use AM-HM Inequality for both.
$$\frac{a+b}{2}\ge \frac{2}{\frac{1}{a}+\frac{1}{b}} \Rightarrow \frac{1}{a}+\frac{1}{b}\ge \frac{4}{a+b}$$
Similarly, you get $$\frac{1}{a}+\frac{1}{c}\ge \frac{4}{a+c}$$ and $$\frac{1}{c}+\frac{1}{b}\ge \frac{4}{c+b}$$
Now add the three and get the left inequality.
For right inequality, $$\frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}{3}\ge \frac{3}{\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}}$$ or 
$$\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\ge \frac{9}{a+b+c}$$
A: The Inequality $$2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq \frac{9}{a+b+c}$$ follows by Cauchy Schwarts. Indeed, multiply by $a+b+c$  and write $$[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\geq 9$$
A: Hints: For the first inequality, note that $\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$ because this is equivalent to $(a+b)^2\geq 4ab$. For the second, apply AM-GM to
$$
(a+b) + (b+c) + (c+a)\quad\text{and}\quad\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}
$$
and then multiply the resulting two inequalities.
A: 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:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{9}{x+y+z} \Rightarrow 3+(\frac{y}{x}+\frac{x}{y}) +(\frac{y}{z}+\frac{z}{y})+ (\frac{z}{x}+\frac{x}{z}) \geq 3+2+2+2=9$
A: Yes, you  can!  By AM-GM we obtain:
$$\sum_{cyc}\frac{2}{a+b}\leq\sum_{cyc}\frac{1}{\sqrt{ab}}\leq\sum_{cyc}\frac{\frac{1}{a}+\frac{1}{b}}{2}=\sum_{cyc}\frac{1}{a}.$$
The right inequality we can prove also by AM-GM:
$$2\sum_{cyc}\frac{1}{a+b}=\frac{1}{a+b+c}\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq$$
$$\geq\frac{1}{a+b+c}\cdot3\sqrt[3]{\prod_{cyc}(a+b)}\cdot\frac{3}{\sqrt[3]{\prod\limits_{cyc}(a+b)}}=\frac{9}{a+b+c}.$$
