${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$ 
$${\sqrt{a+b+c} + \sqrt{a}\over b+c}  + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}$$

I tried AM-GM, which only gives large terms without any answer.
$${\sqrt{a+b+c} + \sqrt{a}\over b+c}  + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge 3\sqrt[3]{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)}$$
Now this will never simplify to anything.
using a different approach i got, 
$${\sqrt{a+b+c} + \sqrt{a}\over b+c}  + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge \sqrt{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right)}$$
which also does not simplify further.
breaking the individual terms on the LHS also does not help nor does multiplying each term on LHS by its conjugate. :<
This was a introductory problem, so i guess there must exist a easy solution which i can't find. 
Any hints will be helpful.
 A: Hint:
If you multiply $a$, $b$ and $C$ by a same positive constant, $k^2$, then both LHS and RHS get multiplied by $k^{-1}$ (i.e. the inequality is homogeneous). Therefore you can assume WLOG that $a+b+c= 1$.
Simplify the inequality and you get:
$$\dfrac{1}{1-\sqrt{a}} + \dfrac{1}{1-\sqrt{b}} + \dfrac{1}{1-\sqrt{c}} \geq \dfrac{9 - 3\sqrt{3}}{2}.$$
Perhaps this is easier to prove.
A: It must be 
$$\sum\limits_{cyc}\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}\geq\frac{9+3\sqrt3}{2\sqrt{a+b+c}}$$
Let $a+b+c=3$.
Hence, 
$$\sum\limits_{cyc}\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}-\frac{9+3\sqrt3}{2\sqrt3}=\sum\limits_{cyc}\frac{\sqrt3+\sqrt{a}}{3-a}-\frac{3\sqrt3+3}{2}=$$
$$=\sum\limits_{cyc}\left(\frac{1}{\sqrt3-\sqrt{a}}-\frac{1}{\sqrt3-1}\right)=\frac{1}{\sqrt3-1}\sum\limits_{cyc}\frac{\sqrt{a}-1}{\sqrt3-\sqrt{a}}=$$
$$=\frac{1}{\sqrt3-1}\sum\limits_{cyc}\left(\frac{\sqrt{a}-1}{\sqrt3-\sqrt{a}}-\frac{1}{2(\sqrt3-1)}(a-1)\right)=$$
$$=\frac{1}{2(\sqrt3-1)^2}\sum\limits_{cyc}\frac{(\sqrt{a}-1)^2(\sqrt{a}+2-\sqrt3)}{\sqrt3-\sqrt{a}}\geq0$$
