<s>[INST] 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?
 [/INST] For $a=b=c=\frac{1}{\sqrt3}$ we obtain a value $\sqrt3$.
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{ab}{c}\geq\sqrt{3(a^2+b^2+c^2)}$$ or
$$\sum_{cyc}a^2b^2\geq\sqrt{3a^2b^2c^2(a^2+b^2+c^2)}$$ or
$$\sum_{cyc}(a^4b^4-a^4b^2c^2)\geq0$$ or
$$\sum_{cyc}c^4(a^2-b^2)^2\geq0$$ and we are done!
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