Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$ 
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
  $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$

I tried various methods. But, couldn't solve it. It'd be great if anyone can help. 

You already have a proof using AM-GM and Muirhead.  For a pure AM-GM proof, note that $\frac12 \left(\frac{a}b+\frac{b}a\right) \geqslant 1$ and so on, so half the LHS takes care of the $3$ on RHS. 
For the rest, note you can cyclically sum the (weighted) AM-GMs:
$$\frac5{18} \left(\frac{a}b+\frac{a}c\right)+\frac2{18} \left(\frac{b}a+\frac{b}c\right)+\frac2{18} \left(\frac{c}a+\frac{c}b\right)\geqslant \frac{\sqrt[3]a}{\sqrt[6]{b c}}=\sqrt{a}$$
