<s>[INST] 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.?
 [/INST] 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.
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