Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2$. Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that 
$$a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2.$$
In my opinion, I do this problem by th following:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(c+a)},$$
but I do not know how to continue following this way.
It seems easy to enlarge this quantity, but how to reduce this quantity.
Any help and hint will welcome!!

I think your idea works, let's proceed:
Introduce the following substitution: $$a + b = u^2$$
$$b + c = v^2$$
$$c + a = w^2$$
Then we have:
$$2a = u^2 + w^2 - v^2$$
$$2b = v^2 + u^2 - w^2$$
$$2c = w^2 + v^2 - u^2$$
$ab + bc + ca = 1$ becomes:
$$4 = 2u^2v^2 + 2v^2w^2 + 2w^2u^2 - u^4 - v^4 - w^4$$
So the inequality becomes:
$$\sum_{cyc}{(u^2 + w^2 - v^2)uw} \geq 2\sum_{cyc}{u^2v^2} - \sum_{cyc}u^4$$
In Muirhead-like notation:
$$T[3, 1, 0] + \frac{1}{2}T[4, 0, 0] \geq T[2, 2, 0] + \frac{1}{2}T[2, 1, 1]$$
This is just pure Muirhead with $T[3, 1, 0] \geq T[2, 2, 0]$ and $T[4, 0, 0] \geq T[2,1,1]$.
