Show that the inequality $\frac{s}{t} + \frac{t}{r} + \frac{r}{s} >= 3$, assuming s,t,r >0.

Question

For this question, I made an assumption that $r<=s<=t$. If I manage to proof it under this assumption, is the result still hold true when $r<=t<=s$ or $ t<=s<=r$, if it is true, why?

Sorry, I didn't learn about AM-GM inequality. Would you mind to explain how AM-GM inequality works in this question here? My knowledge of AM-GM is limited to $a^2+b^2>=2ab $....

Thank you very much for your reply. This question has been confusing me for a long period of time.

Answer 1

$$\sum_{cyc}\left(\frac{s}{t}-\frac{s}{r}\right)=\frac{\sum\limits_{cyc}(s^2r-s^2t)}{str}=\frac{(s-t)(t-r)(r-s)}{str}\geq0$$ for $r\leq s\leq t$, which says that it's enough to prove our inequality for $r\leq t\leq s$.

But it's better to use AM-GM: $$\sum_{cyc}\frac{s}{t}\geq3\sqrt[3]{\frac{s}{t}\cdot\frac{t}{r}\cdot\frac{r}{s}}=3.$$

I use AM-GM for three numbers:

for positives $x$, $y$ and $z$ we have: $$\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$$

or $$x+y+z\geq3\sqrt[3]{xyz}.$$ Now, take $x=\frac{s}{t},$ $y=\frac{t}{r}$ and $z=\frac{r}{s}.$