How can I prove this inequality using HM-GM-AM-QM inequalities? Given $a,b,c\in\mathbb{R}^{+}$, prove that $$(a+b+c)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}\right) \geq \frac{9}{2}$$
I've been trying for a couple of hours in total, and I just can't seem to get it to work no matter what I do.
Edit: I also realize this is a general problem-solving issue I have, as it requires me to pick the right option out of thousands of possible paths I could take with the proof. Is there a generalized way to approach these kinds of questions?
 A: The AM-HM inequality states that for $x,y,z\in\Bbb{R}_{>0}$ $$\frac{x+y+z}{3}\geq\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\\\iff(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq9$$ Now take $x=(a+b),y=(b+c)$ and $z=(c+a)$
One More Proof Using the Famous Nesbitt's Inequality:
$$(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\\=\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\\=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge3+\frac{3}{2}=\frac{9}{2}$$
The last step follows from the very well-known Nesbitt's inequality
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$$
A: Change variables $\alpha=a+b$, $\beta=b+c$ and $\gamma=c+a$ to transform the LHS to
$$
\frac{1}{2}(\alpha+\beta+\gamma)\Big(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\Big)\ .
$$
Now, AM-GM gives
$$
\alpha+\beta+\gamma\geq3\sqrt[3]{\alpha\beta\gamma}
$$
and
$$
\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\geq3\sqrt[3]{\frac{1}{\alpha\beta\gamma}}\ .
$$
Now just multiply the inequalities together.
