If $a,b,c\in(0;+\infty)$, prove that $\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$. If $a,b,c\in(0;+\infty)$ and $$\frac{c}{1+a+b}+\frac{a}{1+b+c}+\frac{b}{1+c+a}\ge\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ca}{1+c+a}$$Prove that $$\frac{a^2+b^2+c^2}{ab+bc+ca}+a+b+c+2\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
I know that $$a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}\ge ab+bc+ac$$
So we could prove that $$a+b+c+3\ge2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$$
I.e. $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3$$
By using AM-GM we see that we could prove that: $$2(a+b+c)\le a+b+c+3\Rightarrow a+b+c\le3$$
So this didn't work.

The given condition is equivalent to: 
$$\sum_{cyc}\frac{c-ab}{1+a+b} \ge 0$$
$$\implies \sum_{cyc}\frac{c(1+a+b)-ab}{1+a+b} \ge  \sum_{cyc}\frac{c(a+b)}{1+a+b} \qquad \text{adding to both sides}$$
$$\implies \sum_{cyc}c \ge  \sum_{cyc}\frac{c(a+b)+ab}{1+a+b} = (ab+bc+ca)\sum_{cyc}\frac{1}{1+a+b} \tag{1}$$
By Cauchy-Schwarz, we also have
$$\left(\sum_{cyc}\frac{1}{1+a+b}\right)\left(\sum_{cyc}(c+ca+bc)\right)\ge \left(\sum_{cyc} \sqrt{c} \right)^2 \tag{2} $$
Using this in $(1)$, we have:
$$a+b+c \ge (ab+bc+ca)\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{a+b+c+2(ab+bc+ca)}$$
Cross multiplying and expanding we get, 
$$(a+b+c)^2 + 2(ab+bc+ca)(a+b+c)\ge (ab+bc+ca)(a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca})$$
Expanding the first term, cancelling part of the second term with RHS and dividing throughout by $ab+bc+ca$, we get the result.
