Algebra question : Prove the inequality. 
Let $a , b \ \& \ c$ be positive real numbers satisfying : $$\cfrac{a}{1+b+c} + \cfrac{b}{1+c+a} + \cfrac{c}{1+a+b} \ge \cfrac{ab}{1+a+b} + \cfrac{bc}{1+b+c}+ \cfrac{ca}{1+a+c} $$ 
  Prove that : 
  $$\cfrac{a^2 + b^2 + c^2}{ab + bc + ca} + a + b + c + 2 \ge 2 \left(\sqrt{ab} + \sqrt{bc} + \sqrt{bc} \right) $$

This is what I've tried yet : 
While, I know that this is not anything like appreciable effort. But, I really have tried a lot after this including the logic of all the denominators being positive.
This is the try for the first equation which didn't lead me anywhere. I tried to simplify the second equation also, this is what I got : 

This looked very good to me. But, again my brain is empty with further ideas. 
The ideas I have got till now include : $$\bullet \text{Using AM-GM inequality} : $$

I don't know what to do further. I thought about this for 1 day and then at each break, an idea came in my mind but all were in vain. 
I will appreciate any help regarding this. And if anyone could comment on my ideas and try to explain how to go on from there too (if there is any way) , then it will be great also.
Thanks!

Using the given inequality
\begin{align*}
& a+b+c-(ab+bc+ca)\ge 0\implies \dfrac{a+b+c}{ab+bc+ca}\ge 1\\
& (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\
\implies & \dfrac{(a+b+c)^2}{ab+bc+ca}=\dfrac{a^2+b^2+c^2}{ab+bc+ca}+2\\
\implies & (a+b+c)\left(\dfrac{a+b+c}{ab+bc+ca}+1\right)=\dfrac{a^2+b^2+c^2}{ab+bc+ca} +(a+b+c)+2\\
\implies & 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})≤2(a+b+c)≤\dfrac{a^2+b^2+c^2}{ab+bc+ca}+(a+b+c)+2
\end{align*}
