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.

Here are some thought I have, but no proof (yet):
Both the condition and the inequality to be proven have equality when $(a,b,c)=(x,x,x)$ and when $(a,b,c)=(1,1,4)$ (or permutations). We know that the inequalities of means only have equality when the term we apply them to are equal. Thus, we can never use AM-GM on $a$, $b$ and $c$, because $(1,1,4)$ wouldn't give equality and thus the inequality is to lossy.
If we apply AM-GM on $f(a,b,c)$, $g(a,b,c)$ and $h(a,b,c)$, we want $f(1,1,4)=g(1,1,4)=h(1,1,4)$, $f(1,4,1)=g(1,4,1)=h(1,4,1)$ and the same for $(1,1,4)$. I think it is not easy to find functions that satisfy this (apart from $f=g=h$, but then, you won't get any information from the inequality).
We can thus only use the given inequality and use rewriting of terms. Maybe we can also use Cauchy-Schwarz, because it has not-so-trivial cases for equality.
