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.
Best Answer
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!