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?
Best Answer
Trivially, we have $(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$, so we get $$(x+y+z)^2 \geq 3(xy+yz+xz)$$ by adding to both sides of the equation. Thus by plugging in $x = \frac{ab}{c}$, $y = \frac{bc}{a}$, $z = \frac{ca}{b}$, we get $$\left(\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b}\right)^2 \geq 3(b^2 + c^2 + a^2) = 3$$ and thus $\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} \geq\sqrt{3}$. We attain equality by setting $a=b=c=\frac{\sqrt{3}}{3}$.