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Solution

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Set $S=\textstyle \sum_i x_i$. By Cauchy-Schwarz, \[ \left(\sum_{i}\frac{x_i}{S-x_i}\right)\left(\sum_{i} x_i(S-x_i)\right)\ge S^2, \]so \[ \sum_i \frac{x_i}{S-x_i} \ge \frac{S^2}{\sum_i x_i(S-x_i)} = \frac{S^2}{S^2-\sum_i x_i^2}. \]Lastly, using $n\textstyle \sum_i x_i^2 \ge S^2$ by Cauchy-Schwarz, we arrive at the desired conclusion.