Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality 
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$ 
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ etc.
Using that I get this
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}}$ 
From here, I used the fact that $(a+b)(b+c)(a+c)\geq 8abc$, which can be easily proven by considering that $a+b\geq 2\sqrt{ab}$ 
But by using this, I get the following...
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq 3\times \sqrt[3]{\dfrac{abc}{(a+b)(b+c)(a+c)}} \leq 3 \times \sqrt[3]{\dfrac{abc}{8abc}} = \dfrac{3}{2}$ 
Everything seems so perfect because I get the value $\dfrac{3}{2}$ as required, but this method isn't valid due to the change in direction! What is going on?
Is there a way of proving this inequality otherwise then?

Here is another nice proof.
We have to prove;
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}\\\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\geq \frac{3}{2}+3\\(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq \frac{9}{2}\\([b+c]+[a+c]+[a+b])\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9\\$$Now we use this well known inequality for $x>0\to x+\frac{1}{x}\geq2$;
(quick proof: $\left (\sqrt x-\frac{1}{\sqrt x}\right )^2\geq0\to x+\frac1x\geq2$)
$$1+1+1+\left[\frac{a+b}{a+c}+\frac{a+c}{a+b}\right]+\left[\frac{a+b}{b+c}+\frac{b+c}{a+b}\right]+\left[\frac{a+c}{b+c}+\frac{b+c}{a+c}\right]\geq 9\\q.e.d.$$
