If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$? 
Suppose that $a^2 + b^2 + c^2 = 1$ for real positive numbers $a$, $b$, $c$. Find the minimum possible value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$.

So far I've got a minimum of $\sqrt {3}$. Can anyone confirm this? However, I've been having trouble actually proofing that this is the lower bound. Typically, I've solved problems where I need to prove an inequality as true, but this problem is a bit different asking for the minimum of an inequality instead, and I'm not sure how to show that $\sqrt {3}$ is the lower bound of it. Any ideas?

We have
$$\left(\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}\right)^2 = \frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2} + \frac {c^2a^2}{b^2}+2(a^2+b^2+c^2).$$
Using the AM-GM inequality, we get
$$\frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2} + \frac {c^2a^2}{b^2} = \frac{1}{2} \sum \left(\frac {a^2b^2}{c^2} + \frac {b^2c^2}{a^2}\right) \geqslant \sum \sqrt{\frac {a^2b^2}{c^2} \cdot  \frac {b^2c^2}{a^2}}=a^2+b^2+c^2.$$
Therefore
$$\left(\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}\right)^2 \geqslant 3(a^2+b^2+c^2) = 3,$$
or
$$\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b} \geqslant \sqrt 3.$$
Equality occur when $a=b=c=\frac{1}{\sqrt{3}}.$
