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
 A: It's just comment.
I think, this inequality is very interesting.
The following way does not help.
By using the Ji Chen's lemma: https://artofproblemsolving.com/community/c6h194103
it's enough to prove three inequalities:

*

*$$\sum_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\geq2+\frac{a^2+b^2+c^2}{ab+ac+bc}$$
2.$$\sum_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\left(\frac{a}{b+c}+\frac{c}{a+b}\right)\geq1+\frac{2(a^2+b^2+c^2)}{ab+ac+bc}$$ and
3.$$\prod_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\geq\frac{a^2+b^2+c^2}{ab+ac+bc}.$$
The first it's just $$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2)\geq0,$$ which is true by Muirhed.

The second is true by Muirhead again:
$$\sum_{sym}\left(a^7b-a^5b^2+a^5b^2c-a^4b^3c+\frac{1}{2}a^6bc-\frac{1}{2}a^3b^3c^2\right)\geq0,$$ but the third is wrong!
It's equivalent to:
$$-abc\sum_{sym}(a^4b-a^3b^2)\geq0.$$
A: Here is a solution with some Matlab help for the analysis, but with a clear manual proof path.
Due to homogeneity, we can demand $a^2+ b^2 + c^2 = 1$. Define $m$ to be the mean of $a,b,c$, i.e. $a + b+c = 3m$. Then note that
$$
9 m^2 = (a+b+c)^2 = a^2+ b^2 + c^2 + 2 (ab + bc + ca) = 1 +  2(ab + bc + ca)
$$
Hence the claim can be written
$$
\sum_{cyc} \sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\ge 2+\sqrt{\frac{2}{9m^2-1}}
$$
Now turn to the LHS. Write $a = m +x$, $b = m + y$, $c = m+z$ with $x+y+z=0$ and $1 = a^2 + b^2 + c^2 = 3 m^2 + x^2 + y^2 + z^2$ which gives two conditions for $(x,y,z)$. W.l.o.g. $(x,y,z)$ can then be expressed as
$$
x =  \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi-2\pi/3)\\
y =  \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi-4\pi/3)\\
z =  \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi)
$$
Hence the claim can be written, with these $(x,y,z)$, as
$$
\sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}\ge 2+\sqrt{\frac{2}{9m^2-1}}
$$
The LHS is now a function of $\phi$ whereas the RHS is not. For any $m$, a free (unbounded) minimum w.r.t. $\phi$ of the LHS occurs at $\phi = \pi$ which can be shown by varying $\phi$ about $\pi$. [For bounded minima see below.] So we have to inspect the LHS at that minimum and show that
$$
\lim_{(\phi = \pi)} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}-  2-\sqrt{\frac{2}{9m^2-1}} \ge 0
$$
Since $(a,b,c)$ should be nonnegative, this requires that $c = m + z = m  - \sqrt\frac23 \sqrt{1 - 3m^2} > 0$ or $m > \sqrt2 / 3$, this bound corresponds to $(a,b,c) = (\frac{1}{\sqrt2},\frac{1}{\sqrt2},0)$.  On the other hand, the maximum possible $m$ occurs when $a = b = c = m$ or, since $a^2+b^2 + c^2 = 1$, at $m = 1/\sqrt3$.
Let's look at the two extreme values for $m$. Indeed we have (using Matlab) that
$$
\lim_{(m = \sqrt2 / 3)} \lim_{(\phi = \pi)} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}-  2-\sqrt{\frac{2}{9m^2-1}} = 0 \\
\lim_{(m = 1 / \sqrt3)} \lim_{(\phi = \pi)} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}-  2-\sqrt{\frac{2}{9m^2-1}} = 0 
$$
and for all values of $m$ in between the $> 0 $ holds. Below is a plot which illustrates this.

The minimum of the LHS may as well be bounded by the fact that $(a,b,c)$ should be nonnegative. In that case, the bound arises when the smallest variable, say $c$, is zero, and it must be inspected, while keeping $c=0$, until another variable becomes zero. So that bound is given by $0 = c = m + \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi)$ or $m = \sqrt{\frac{\frac23 \cos^2(\phi) }{1 + 2 \cos^2(\phi) }}$ and must be inspected for $\frac43 \pi > \phi > \frac23 \pi$ since at $\frac23 \pi$ (or $\frac43 \pi$ ) we have that also $b =0$ or $a =0$ (then the terms diverge, and this case was excluded by the OP).   That means we have to look at (with the $(x,y,z)$ as above)
$$
\lim_{m = \sqrt{\frac{\frac23 \cos^2(\phi) }{1 + 2 \cos^2(\phi) }}}  \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}-  2-\sqrt{\frac{2}{9m^2-1}}
$$
which is a function of $\phi$. Variation of $\phi$ about $\pi$ already  shows local positivity. Here is a plot (where $\phi$ was denoted $x$) which illustrates the overall behavior:

This proves the claim. $\qquad \Box$
A: 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 $z(k+1)+xy=1$
