Solution:
First we will prove a simple lemma involving only two variables: For all positive $a, b$,
$$\left(a^{2}+1\right)\left(b^{2}+1\right) \geq (a b+1)^{2}$$
To see why this is true, multiply out, and after simplifying, we have
$$a^{2}+b^{2} \geq 2 a b$$
This is equivalent to
$$a^{2}-2 a b+b^{2}=(a-b)^{2} \geq 0$$
which of course is true (in fact, for any real numbers $a$ and $b$).

Now we shall attack the problem. Multiplying both sides by $x_{1} x_{2} \cdots x_{n}$ produces the equivalent inequality
$$\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right) \cdots\left(x_{n}^{2}+1\right) \geq\left(x_{1} x_{2}+1\right)\left(x_{2} x_{3}+1\right) \cdots\left(x_{n} x_{1}+1\right) .$$
Applying the lemma repeatedly yields
$$\begin{aligned} &\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+1\right) \geq\left(x_{1} x_{2}+1\right)^{2}, \\ &\left(x_{2}^{2}+1\right)\left(x_{3}^{2}+1\right) \geq\left(x_{2} x_{3}+1\right)^{2}, \\ & \vdots \\ &\left(x_{n}^{2}+1\right)\left(x_{1}^{2}+1\right) \geq\left(x_{n} x_{1}+1\right)^{2} . \end{aligned}$$
Multiplying these yields the square of the desired inequality.

Solution:
We shall use induction. Even though the problem begins with $n=2$, we can start by noting that for $n=1$, the statement is merely the trivial
$$x_{1}+1 / x_{1} \geq x_{1}+1 / x_{1}$$
In general, suppose without loss of generality that $x_{1}$ is the largest among the given $n$ numbers. The right-hand side products containing $x_{1}$ are: $\left(x_{1}+1 / x_{2}\right)\left(x_{n}+1 / x_{1}\right)$. We claim that this product will not decrease if we swap the places of $x_{2}$ from the first multiple and $x_{1}$ from the second multiple, i.e.
$$\left(x_{1}+1 / x_{2}\right)\left(x_{n}+1 / x_{1}\right) \leq\left(x_{1}+1 / x_{1}\right)\left(x_{n}+1 / x_{2}\right) .$$
This inequality is easy to prove: after a little algebra, it becomes
$$x_{1} x_{2}+x_{1} x_{n} \leq x_{1} x_{1}+x_{2} x_{n}$$
which is equivalent to
$$\left(x_{1}-x_{n}\right)\left(x_{1}-x_{2}\right) \geq 0$$
and this is true because $x_{1}$ was the largest number among the given $n$ numbers.

Notice that after performing the "swap," we may cancel $\left(x_{1}+1 / x_{1}\right)$ from both sides, and what we are left with is the same problem but for the $(n-1)$ numbers $x_{2}, x_{3}, \ldots x_{n}$. This completes the inductive step.