If $a^2 + b^2 + c^2 = 1$, what is the the minimum value of $\frac {ab}{c} + \frac {bc}{a} + \frac {ca}{b}$?

Question

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?

Answer 1

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 $a=b=c=\frac{\sqrt{3}}{3}$.

Answer 2

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 $a=b=c=\frac{1}{\sqrt{3}}.$

Answer 3

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!

Answer 4

I found a stronger version: $$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge \sqrt{4(a^2+b^2+c^2)-ab-bc-ca}$$ is true $\forall a,b,c>0$

Note: It's just Schur 3 degree