<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] We will prove the homogeneous inequality (where we used $abc=1$) $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[6]{abc}$$ Since equality is attainable for $a=b=c$, this would show that the desired $k$ is $3$. Observe that, in virtue of AM-GM $$\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{b}}+\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{c}}}6\geqslant \sqrt[6]{abc}\iff\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{b}}+\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{c}}}2\geqslant 3\cdot \sqrt[6]{abc} $$
Also, AM-GM yields $$\frac{2\cdot\frac{a}{\sqrt{b}}+2\cdot \frac{a}{\sqrt{c}}+\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}}{6}\geqslant \sqrt[6]{a^3}=\sqrt{a}$$ Adding the first inequality and the cyclic variations of the latter one, we arrive at the desired inequality.
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