Inequality proof $\frac{x}{y}+2\frac{y}{z}+3\frac{z}{x} \geq 3\sqrt[3]{6} $

Question

I have found numerically that for $x>0,y>0,z>0$ we have $$ \frac{x}{y}+2\frac{y}{z}+3\frac{z}{x} \geq 3\sqrt[3]{6} $$ but I don't know how to prove this. Do you have any ideas?

Answer 1

It's quite interesting you found it numerically. How did you do it?

However you can prove your observation using the lemma $$a+b+c \ge 3 \sqrt[3]{abc}$$ For $a=\dfrac{x}{y}$ , $b=\dfrac{2y}{z}$ and $c=\dfrac{3z}{x}$ the prove follows.