Problem 1492

1. Proof: When $n=1$, the conclusion is obviously true. Now assume that the conclusion holds for $n=1, \cdots, l(l \geqslant 1)$. Let's consider the case when $n=l+1$. That is, there are $l+1$ non-negative real numbers $x_{1}, x_{2}, \cdots, x_{l+1}$ satisfying $x_{1} x_{2} \cdots x_{l+1}=1$. We will discuss the following cases:

Case one. At least one $x_{i}=1$, for example, let $x_{l+1}=1$, then we have $x_{1} \cdots x_{l}=x_{1} \cdots x_{l} x_{l+1}=1$. By the induction hypothesis, we get $x_{1}+\cdots+x_{l} \geqslant l$, thus $x_{1}+\cdots+x_{l}+x_{l+1} \geqslant l+1$.

Case two. Assume $x_{1}, \cdots, x_{l+1}$ are all not equal to 1. From $x_{1} \cdots x_{l+1}=1$, it is easy to see that it is impossible for $x_{1}, \cdots, x_{l+1}$ to all be less than 1 or all greater than 1. Therefore, we can assume without loss of generality that $x_{l} < 1$ and $x_{l+1} > 1$. Applying the induction hypothesis to $x_{1}, \cdots, (x_{l} x_{l+1})$, we get
$$x_{1}+x_{2}+\cdots+(x_{l} x_{l+1}) \geqslant l$$

From $x_{l} < 1$ and $x_{l+1} > 1$, it is easy to see that
$$x_{l} x_{l+1} > 1$$

This proves that the conclusion holds for $n=l+1$. Therefore, the inequality holds for any natural number $n$.