Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$

Question

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

Answer 1

By AM-GM $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge 2(\sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} )=2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$ Now using $x^2+y^2+z^2\ge xy+yz+zx$ $$\begin{align*}2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge 2(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}})=2(\sqrt{a}+\sqrt{b}+\sqrt{c})& \\ \ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}\end{align*}$$ as $abc=1$ we see that $k=3$

Equality when $a=b=c=1$

Answer 2

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.

Answer 3

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.