<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] 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 $$[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\geq 9$$
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