<s>[INST] If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{a+b+c+15}}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}$$
It seems nice enough.
I proved this inequality by Holder, but it quits very ugly.
Maybe there is something nice? Thank you!
 [/INST] $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq\frac{9}{\sqrt{a+b+c+15}}\iff\sqrt{a+b+c+15}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\right)\ge9$$
Making $c=\frac{1 } {ab}$ the expression becomes
$$f(a,b)= \left(\sqrt{\frac{a^2b}{ab^2+1}}+\sqrt{\frac{ab^2}{a^2b+1}}+\sqrt{\frac{1}{ab(a+b)}}\right)\sqrt{\frac{ab(a+b+15)+1}{ab}}\ge9$$ for all positive $a,b$.
It follows $$\sqrt{\frac{a^3b+a^2b^2+15a^2b+a}{ab^2+1}}+\sqrt{\frac{ab^3+a^2b^2+15ab^2+b}{a^2b+1}}+\frac{1}{ab}\sqrt{\frac{ab(a+b+15)+1}{a+b}} \ge9$$
It is clear $f(x,y)$ has no maximum and, in order to prove the inequality, we want to get the minimum of $f(x,y)$. 
This minimum can be calculated as usually for two variables ($f_x(x,y)=0$ and $f_y(x,y)=0$,  etc). 
We calculate as follows:  since $f(a,b)=f(b,a)$ the minimum of $f(a,b)$ is equal to the minimum of $f(a,a)$ where $a\gt 0$. Hence we calculate the minimum of the function of one variable
$$f(x,x)=2\sqrt{\frac{2x^4+15x^3+x}{x^3+1}}+\frac{1}{x^2}\sqrt{\frac{2x^3+15x^2+1}{2x}}$$
The calculation is straightforward although somewhat tedious giving the minimum $9$ for $x=1$. For further explanation, see figure below wherein the calculation (Wolfram) and the graph of the function (Desmos) confirm the result. Thus this minimum is attained with $a = b = c = 1$ and the proposed inequality is valid for all positive with $abc=1$.

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