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

Let $y_k = \dfrac{m}{x_k + m}, 0 < y_k <1, \sum{y_k}=1$. Since $x_k=m \cdot \left(\dfrac{1-y_k}{y_k}\right)$, the inequality to be proven translates as $\prod{(1-y_k)}\ge (n-1)^n\prod{y_k}$.

Noting that $1-y_k =\sum {y_q}-y_k$, applying the mean inequality gives $(1-y_k)^{n-1} \ge (n-1)^{n-1}\dfrac{\prod {y_q}}{y_k}$ for eack $k=\overline{1, n}$.

Multiplying all these inequalities and taking the root of order $\dfrac{1}{n-1}$ at the end we are done!

$$\sum_{k\neq i}\frac{1}{x_k+m}=\frac{1}{m}-\frac{1}{x_i+m}=\frac{x_i}{m(x_i+m)}.$$ Thus, by AM-GM $$\prod_{i=1}^n\frac{x_i}{x_i+m}=m^n\prod_{i=1}^n\sum_{k\neq i}\frac{1}{x_k+m}\geq\frac{m^n(n-1)^n}{\prod\limits_{i=1}^n\left(\prod\limits_{k\neq i}(x_k+m)\right)^{\frac{1}{n-1}}}=\frac{m^n(n-1)^n}{\prod\limits_{i=1}^n(x_i+m)},$$ which ends a proof.