The inequality. Regional olympiad 2015 Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$
Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$
Probably, we should use these facts:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{abc}$
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2$
But I don't know how to use them. Please, help.  

We have 
$$
\sum_{cyc} \sqrt{a+\frac{1}{a}}=\sum_{cyc} \sqrt{a+\frac{ab+bc+ca}{a}}=\sum_{cyc} \sqrt{a+\frac{bc}{a}+b+c}\ge\sum_{cyc} \sqrt{2\sqrt{bc}+b+c}=\sum_{cyc} \sqrt{\left(\sqrt b+\sqrt c\right)^2}=\sum_{cyc} \left(\sqrt b+\sqrt c\right)=2\left(\sqrt a+\sqrt b+\sqrt c\right)
$$
as desired.
Note: The inequality is due to the well known inequality between the arithmetic mean and the geometric mean, which states that:
$$
\frac{a_1+…+a_n}{n}\ge\left(a_1\cdots a_n\right)^\frac{1}{n}
$$
Taking $n=2$, $a_1=a$ and $a_2=\frac{bc}{a}$ this yields:
$$
\frac{a+\frac{bc}{a}}{2}\ge\left(a\cdot\frac{bc}{a}\right)^\frac12=\sqrt{bc}\iff a+\frac{bc}{a}\ge2\sqrt{bc}
$$
This is, as shown above, enough to prove your inequality.
