<s>[INST] 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. 
 [/INST] By AM-GM
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}
\ge \frac{2\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}
= \frac2a+\frac2b+\frac2c$$
For the last equality $abc=1$ was used.
Then again by AM-GM; 
$$\frac1a+\frac1b+\frac1c \ge \frac3{\sqrt[3]{abc}}=3$$
and again by a well known inequality 
$$\frac1a+\frac1b+\frac1c \ge \frac1{\sqrt{bc}}+\frac1{\sqrt{ca}}+\frac1{\sqrt{ab}}=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
For the last equality $abc=1$ was used. Combine the last two inequalities to get the desired result.
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