<s>[INST] ${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$ 
$${\sqrt{a+b+c} + \sqrt{a}\over b+c}  + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}$$

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.
 [/INST] 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.
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