Prove that $\frac{\sqrt{\prod_{i=1}^n}x_i}{n-1} \geq 1998$

Question

Let $n \geq 2$, $x_1,x_2,x_3....x_n \in R^+$ such that $$\sum_{i=1}^n\frac{1}{x_i+1998} = \frac{1}{1998}.$$ Prove that $$\frac{\sqrt{\prod_{i=1}^n x_i}}{n-1} \geq 1998$$

Answer 1

I'll try from $\sum\limits_{i=1}^n\frac{1}{x_i+1998}=\frac{1}{1998}$.

For all $i$ by AM-GM we have $\frac{1}{1998}-\frac{1}{x_i+1998}=\sum\limits_{k\neq i}\frac{1}{x_k+1998}\geq\frac{n-1}{\sqrt[n-1]{\prod\limits_{k\neq i}(x_k+1998)}}$.

Thus, $\frac{1}{1998^n}\cdot\frac{\prod\limits_{k=1}^nx_k}{\prod\limits_{k=1}^n(x_k+1998)}\geq\frac{(n-1)^n}{\prod\limits_{k=1}^n(x_k+1998)}$, which gives $\sqrt[n]{\prod\limits_{k=1}^nx_k}\geq1998(n-1)$.