<s>[INST] Prove that $\sum_{cyc} \sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\ge 2+\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ For $a,b,c\geq 0$, no two of which are $0$, prove that:
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\dfrac{a^2+b^2+c^2}{ab+bc+ca}}$$
This inequality actually came up as an accident when I tried to combine 2 known results, and after many testings on computer it still remains true, but there's still no original proof yet. Hope everyone enjoy and have some good ideas for it.
Here's that 2 known results:
$$\dfrac{a^2+b^2+c^2}{ab+bc+ca}\geq \prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)$$
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)}$$
The second one can be proved by direct Karamata's inequality, but it may also inspire some ideas for the original one too.
See the following links:
https://artofproblemsolving.com/community/u410204h2218857p16854913
https://artofproblemsolving.com/community/c6h487722p5781880
https://artofproblemsolving.com/community/u414514h2240506p17302184
 [/INST] As Michael Rozenberg it's just a comment .Due to homogeneity we can assume that $a=1$ and $0<b,c\leq 1$ we have :
$$\sqrt{\dfrac{1}{b+c}+\dfrac{b}{c+1}}+\sqrt{\dfrac{b}{c+1}+\dfrac{c}{1+b}}+\sqrt{\dfrac{c}{1+b}+\dfrac{1}{b+c}}\geq 2+\sqrt{\dfrac{b^2+c^2+1}{b+bc+c}}\quad (1)$$
We can also assume that $b+c=k=\operatorname{constant}$ and try the substitution :
$$x=\dfrac{b}{c+1}$$
$$y=\dfrac{c}{1+b}$$
$$z=\dfrac{1}{(1+b)(c+1)}$$
$(1)$ becomes :
$$\sqrt{\dfrac{1}{k}+x}+\sqrt{x+y}+\sqrt{y+\dfrac{1}{k}}\geq 2+\sqrt{(\frac{x}{z}+\frac{y}{z}-k+1)\dfrac{z}{1-z}}$$
With the constraint $z(k+1)+xy=1$
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