Three. (50 points) Let non-negative real numbers $x_{i}(i=1,2, \cdots, n)$ for any $n>2$, have
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{n-1} x_{n}+x_{n} x_{1} \\
\geqslant x_{1}^{a} x_{2}^{b} x_{3}^{a}+x_{2}^{a} x_{3}^{b} x_{4}^{a}+\cdots+x_{n}^{a} x_{1}^{b} x_{2}^{a} .
\end{array}
$$

Find the values of the positive real numbers $a$ and $b$.
When $n=4$, equation (1) becomes
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{1} \\
\geqslant x_{1}^{a} x_{2}^{b} x_{3}^{a}+x_{2}^{a} x_{3}^{b} x_{4}^{a}+x_{3}^{a} x_{4}^{b} x_{1}^{a}+x_{4}^{a} x_{1}^{b} x_{2}^{a} .
\end{array}
$$

Let $x_{1}=x_{2}=x_{3}=x_{4}=x \neq 0$, then from equation (2) we get $x^{2} \geqslant x^{2 a+b}$,
which means $1 \geqslant x^{2 a+b-2}$ for $x \in(0,+\infty)$.
If $x>1$, then $2 a+b-2 \leqslant 0$;
If $0<x<1$, then $x \rightarrow+\infty$ when $x^{2 a-1} \rightarrow+\infty$;
If $2 a-1<0$, then $x \rightarrow 0$ when $x^{2 a-1} \rightarrow+\infty$.
Therefore, $2 a-1=0$, which means $a=\frac{1}{2}, b=1$.
Next, we prove that when $a=\frac{1}{2}, b=1$, equation (1) holds.
The left side of equation (1)
$$
\begin{array}{l}
=\sum_{i=1}^{n} x_{i} x_{i+1}=\frac{1}{2} \sum_{i=1}^{n}\left(x_{i} x_{i+1}+x_{i+1} x_{i+2}\right) \\
\geqslant \frac{1}{2} \sum_{i=1}^{n} 2 x_{i}^{\frac{1}{2}} x_{i+1} x_{i+2}^{\frac{1}{2}}=\sum_{i=1}^{n} x_{i}^{\frac{1}{2}} x_{i+1} x_{i+2}^{\frac{1}{2}}
\end{array}
$$
$=$ the right side of equation (1).
Where, $x_{n+k}=x_{k}$ and when $n$ is odd, the equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}$; when $n$ is even, the equality holds if and only if $x_{1}=x_{3}=\cdots=x_{n-1}, x_{2}=x_{4}=\cdots=x_{n}$.
In conclusion, the value of $(a, b)$ is $\left(\frac{1}{2}, 1\right)$.