Let $n\geq 3$ be an integer and let $ x_1,x_2,.......,x_n$ be positive numbers such that $$\sum_{j=1}^n \frac{1}{(1+x_j)}=1$$ Prove that $$\sum_{j=1}^n\sqrt{x_j} \geq (n-1)\sum_{j=1}^n \frac {1}{\sqrt{x_j}}$$ My Approach: Without loss of generality we may assume $x_1\geq x_2\geq x_3 \geq..... \geq x_n$. Then $ \sqrt{x_j}$ and, $\frac{x_j}{1+x_j}$ are increasing sequences ; $\sqrt{\frac{1}{x_j}}$ and, $\frac{1}{1+x_j}$ are decreasing secquences.
Then I applied Chebyshev's Inequality on sequences $ \sqrt{x_j}$ and, $\frac{1}{1+x_j}$ and got the following result $$\sum_{j=1}^n \sqrt {x_j} \geq n\sum_{i=1}^n \frac{\sqrt {x_j}}{(1+x_j)}$$ Further applying Chebyshev's Inequality on sequences $\sqrt{\frac{1}{x_j}}$ and,$\frac{x_j}{1+x_j}$ I got $$(n-1)\sum_{j=1}^n \frac{1}{\sqrt x_j} \geq n\sum_{i=1}^n \frac{\sqrt {x_j}}{(1+x_j)}$$ Did I write anything wrong or write any inequality sign wrong? Any hint shall be much appreciated.
