HMMT February, 2022

Solution:
We will use induction. The base case is $n=3$. In this case, we want to show that
$$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\frac{a_{3}}{a_{1}}>\frac{a_{2}}{a_{1}}+\frac{a_{3}}{a_{2}}+\frac{a_{1}}{a_{3}}.$$
Equivalently, we want to show
$$\begin{aligned} a_{1}^{2} a_{3}+a_{2}^{2} a_{1}+a_{3}^{2} a_{2}>a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+a_{3}^{2} a_{1} & \Longleftrightarrow a_{1}^{2}\left(a_{3}-a_{2}\right)+a_{3}^{2}\left(a_{2}-a_{1}\right)>a_{2}^{2}\left(a_{3}-a_{1}\right) \\ & \Longleftrightarrow\left(a_{3}^{2}-a_{2}^{2}\right)\left(a_{2}-a_{1}\right)>\left(a_{2}^{2}-a_{1}^{2}\right)\left(a_{3}-a_{2}\right) \\ & \Longleftrightarrow a_{3}+a_{2}>a_{2}+a_{1}, \end{aligned}$$
which is true.

Now assume the claim is true for $n \geq 3$. Then, we have that
$$\frac{a_{1}}{a_{3}}+\frac{a_{3}}{a_{4}}+\cdots+\frac{a_{n}}{a_{n+1}}+\frac{a_{n+1}}{a_{1}}>\frac{a_{3}}{a_{1}}+\frac{a_{4}}{a_{3}}+\cdots+\frac{a_{n+1}}{a_{n}}+\frac{a_{1}}{a_{n+1}}.$$
We also have that
$$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\frac{a_{3}}{a_{1}}>\frac{a_{2}}{a_{1}}+\frac{a_{3}}{a_{2}}+\frac{a_{3}}{a_{1}}.$$
Adding the two inequalities and simplifying gives the desired result.

Solution:
The points $\left(a_{1}, \frac{1}{a_{1}}\right),\left(a_{2}, \frac{1}{a_{2}}\right),\left(a_{3}, \frac{1}{a_{3}}\right), \ldots,\left(a_{n}, \frac{1}{a_{n}}\right)$ form a counter-clockwise oriented polygon. Thus, we have the area, $A$, which must be positive, can be calculated by Shoelace theorem:
$$A=\frac{1}{2}\left(\sum_{k=1}^{n} \frac{a_{k}}{a_{k+1}}-\sum_{k=1}^{n} \frac{a_{k+1}}{a_{k}}\right).$$
Since $A$ is positive, we are done.

Solution:
For $1 \leq i \leq n-1$, let $r_{i}=a_{i+1} / a_{i}$. Then the inequality becomes
$$r_{1} r_{2} \cdots r_{n-1}+\frac{1}{r_{1}}+\cdots+\frac{1}{r_{n-1}}>\frac{1}{r_{1} r_{2} \cdots r_{n-1}}+r_{1}+\cdots+r_{n-1}.$$
If we let $s_{i}=\log r_{i}$ and $f(s)=e^{s}-e^{-s}$, this is the same as
$$f\left(s_{1}+\cdots+s_{n-1}\right)>f\left(s_{1}\right)+\cdots+f\left(s_{n-1}\right)$$
This follows from the convexity of $f$ and the fact that $f(0)=0$.