<s>[INST] Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $
How to go further
 [/INST] An alternative way: By the rearrangement inequality we have $$\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\geq \frac{a}{\sqrt{a}}+\frac{a}{\sqrt{a}}+\frac{b}{\sqrt{b}}+\frac{b}{\sqrt{b}}+\frac{c}{\sqrt{c}}+\frac{c}{\sqrt{c}}=2(\sqrt{a}+\sqrt{b}+\sqrt{c})$$ Now proceed by AM-GM as in the answer by Albus Dumbledore.
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