<s>[INST] 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!!
 [/INST] By C-S, Schur and AM-GM we obtain:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(a+c)}\geq\sum_{cyc}a(a+\sqrt{bc})=$$
$$=\sum_{cyc}(a^2-\sqrt{a^3b}-\sqrt{a^3c}+a\sqrt{bc})+\sum_{cyc}(\sqrt{a^3b}+\sqrt{ab^3})\geq2\sum_{cyc}ab=2.$$
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