<s>[INST] 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.  
 [/INST] By Cauchy-Schwarz we have
$$
\sqrt{a + \frac{1}{a}} = \sqrt{a + b + c + \frac{bc}{a}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{bc}{a}}\right)
$$
and similarly
$$
\sqrt{b + \frac{1}{b}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ca}{b}}\right)
$$
and
$$
\sqrt{c + \frac{1}{c}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ab}{c}}\right).
$$
Thus
$$
\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} + \sqrt{c + \frac{1}{c}} 
\geq \frac{1}{2}\left(3\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right) + \frac{\sqrt{a}\sqrt{b}}{\sqrt{c}} + \frac{\sqrt{b}\sqrt{c}}{\sqrt{a}} + \frac{\sqrt{c}\sqrt{a}}{\sqrt{b}} \right) \geq 2\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right).
$$
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