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?

By Cauchy-Schwarz $$\begin{align}\sum_{cyc}\frac{a}{b+c}&=\sum_{cyc}\left(\frac{a}{b+c}+1\right)-3\\
&=\sum_{cyc}a\sum_{cyc}\frac{1}{b+c}-3\\
&=\frac{1}{2}\sum_{cyc}(b+c)\sum_{cyc}\frac{1}{b+c}-3\\
&\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.\end{align}$$
Done!
Another way.
By C-S $$\sum\limits_{cyc}\frac{a}{b+c}=\sum\limits_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$
where the last inequality it's just $\sum\limits_{cyc}(a-b)^2\geq0$.
Done!
Another way.
$$\sum\limits_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits_{cyc}\frac{2a-b-c}{2(b+c)}=\sum\limits_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$
$$=\sum\limits_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum\limits_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Done!
