For which positive real numbers $a, b$ does the inequality

$$
x_{1} \cdot x_{2}+x_{2} \cdot x_{3}+\cdots+x_{n-1} \cdot x_{n}+x_{n} \cdot x_{1} \geq x_{1}^{a} \cdot x_{2}^{b} \cdot x_{3}^{a}+x_{2}^{a} \cdot x_{3}^{b} \cdot x_{4}^{a}+\cdots+x_{n}^{a} \cdot x_{1}^{b} \cdot x_{2}^{a}
$$

hold for all integers $n>2$ and positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$?
Substituting $x_{i}=x$ easily yields that $2 a+b=2$. Now take $n=4, x_{1}=x_{3}=x$ and $x_{2}=x_{4}=1$. This gives $2 x \geq x^{2 a}+x^{b}$. But the inequality between the arithmetic and geometric mean yields $x^{2 a}+x^{b} \geq$ $2 \sqrt{x^{2 a} x^{b}}=2 x$. Here equality must hold, and this implies that $x^{2 a}=x^{b}$, which gives $2 a=b=1$.

On the other hand, if $b=1$ and $a=\frac{1}{2}$, we let $y_{i}=\sqrt{x_{i} x_{i+1}}$ for $1 \leq i \leq n$, with $x_{n+1}=x_{1}$. The inequality then takes the form

$$
y_{1}^{2}+\cdots+y_{n}^{2} \geq y_{1} y_{2}+y_{2} y_{3}+\cdots+y_{n} y_{1} \text {. }
$$

But the inequality between the arithmetic and geometric mean yields

$$
\frac{1}{2}\left(y_{i}^{2}+y_{i+1}^{2}\right) \geq y_{i} y_{i+1}, \quad 1 \leq i \leq n,
$$

where $y_{n+1}=y_{n}$. Adding these $n$ inequalities yields the inequality (1).

The inequality (1) can also be obtained from the Cauchy-Schwarz inequality, which implies that $\sum_{i=1}^{n} y_{i}^{2} \sum_{i=1}^{n} y_{i+1}^{2} \geq\left(\sum_{i=1}^{n} y_{i} y_{i+1}\right)^{2}$, which is exactly the stated inequality.