Prove that $\frac{\sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$ where $\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$.

Given positives $x_1, x_2, \cdots, x_{n - 1}, x_n$ such that $$\large \sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m}$$. Prove that $$\large \frac{\displaystyle \sqrt[n]{\prod_{k = 1}^nx_n}}{m} \ge n - 1$$

We have that $$\sum_{k = 1}^n\frac{1}{x_k + m} = \frac{1}{m} \iff \sum_{k = 1}^n\frac{m}{x_k + m} = 1 \iff \sum_{k = 1}^n\frac{x_k}{x_k + m} = n - 1$$

Let $x_1 \ge x_2 \ge \cdots \ge x_{n - 1} \ge x_m$, using the Hölder's inequality, it is seen that $$\left(\sum_{k = 1}^nx_k\right) \cdot \left(\sum_{k = 1}^n\frac{1}{x_k + m}\right) \ge m \cdot \sum_{k = 1}^n\frac{x_k}{x_k + m} \implies \left(\sum_{k = 1}^nx_k\right) \cdot \frac{1}{m} \ge m(n - 1)$$

Unfortunately, I can't go for more.

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