[Proof] When $n=1$, the inequality obviously holds.
Below, we prove by mathematical induction that the inequality holds for $n=2^{k}$ (where $k$ is a non-negative integer). Assume that the inequality holds for $n=2^{m}$ (where $m$ is some non-negative integer). If $r_{1}, r_{2}, \cdots, r_{2 n} \geqslant 1$, then we have
$$\begin{aligned} \sum_{j=1}^{2 n} \frac{1}{r_{j}+1} & =\sum_{j=1}^{n} \frac{1}{r_{j}+1}+\sum_{j=n+1}^{2 n} \frac{1}{r_{j}+1} \\ & \geqslant \frac{n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}+\frac{n}{\sqrt[n]{r_{n+1} r_{n}+2 \cdots r_{2 n}}+1} \\ & \geqslant \frac{2 n}{\sqrt[2 n]{r_{1} r_{2} \cdots r_{2 n}}+1} \end{aligned}$$

Therefore, for any non-negative integer $k$, the inequality holds when $n=2^{k}$.
For any natural number $n$, if $m=2^{k}>n, k \in \mathbb{N}$, then let
$$r_{n+1}=r_{n+2}=\cdots=r_{m}=\sqrt[n]{r_{1} r_{2} \cdots r_{n}}$$

Thus, we have
$$\sum_{j=1}^{m} \frac{1}{r_{j}+1}=\sum_{j=1}^{n} \frac{1}{r_{j}+1}+\frac{m-n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$$

On the other hand, from the proof above, we know that
$$\begin{array}{l} \sum_{j=1}^{m} \frac{1}{r_{j}+1} \geqslant \frac{m}{\sqrt[m]{r_{1} r_{2} \cdots r_{m}}+1} \\ =\frac{m}{\left(r_{1} r_{2} \cdots r_{n}\right)^{\frac{1}{m}}\left(r_{n+1}^{m-n}\right)^{\frac{1}{m}}+1} \\ =\frac{m}{\left(r_{1} r_{2} \cdots r_{n}\right)^{\frac{1}{m}} \cdot\left[\left(r_{1} r_{2} \cdots r_{n}\right)^{\frac{m-n}{n}}\right]^{\frac{1}{m}}+1} \\ =\frac{m}{\left(r_{1} r_{2} \cdots r_{n}\right)^{\frac{1}{n}}+1} \\ =\frac{m}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1} . \end{array}$$

From the above two equations, we have
$$\sum_{j=1}^{n} \frac{1}{r_{j}+1}+\frac{m-n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1} \geqslant \frac{m}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$$

Thus,
$$\sum_{j=1}^{n} \frac{1}{r_{j}+1} \geqslant \frac{n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$$