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Let $x_1,\dots, x_n\in \mathbb{R}_{++}$ such that $\sum_{i=1}^{n}{x_i}=1$. Prove that $$ \sum_{i=1}^{n}{\frac{x_i}{\sqrt{1-x_i}}} \geq \sqrt{\frac{n}{n-1}} $$

I tried using AM-GM and Cauchy-Schwarz but didn't come to anything useful. Hint could be an help too.

Please advise.

Dennis
  • 289

2 Answers2

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The function $f(x)=x/\sqrt{1-x}$ is convex on $(-1,1)$, so by Jensen's inequality, one has \begin{align*} \sum\dfrac{1}{n}\dfrac{x_{i}}{\sqrt{1-x_{i}}}\geq \dfrac{\displaystyle\sum\dfrac{1}{n}x_{i}}{\sqrt{1-\displaystyle\sum\dfrac{1}{n}x_{i}}}. \end{align*}

user284331
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$$ \sum_{i=1}^{n}{\frac{x_i}{\sqrt{1-x_i}}} =\sum_{i=1}^n {\frac{1-(1-x_i)}{\sqrt{1-x_i}}}=\sum_{i=1}^n{\frac{1}{\sqrt{1-x_i}}}-\sum_{i=1}^n{\sqrt{1-x_i}} $$

Using Cauchy-Schwarz, $$ \sum_{i=1}^n{\sqrt{1-x_i}}\le\sqrt{\sum_{i=1}^n{(1-x_i)}\cdot\sum_{i=1}^n{1}}=\sqrt{(n-1)n} $$ then $$ \sum_{i=1}^n{\frac{1}{\sqrt{1-x_i}}} \ge \frac{\left(\sum\limits_{i=1}^n{1}\right)^2}{\sum\limits_{i=1}^n{\sqrt{1-x_i}}} \ge \frac{n^2}{\sqrt{(n-1)n}} $$

therefore $$ \sum_{i=1}^{n}{\frac{x_i}{\sqrt{1-x_i}}} \ge \frac{n^2}{\sqrt{(n-1)n}} -\sqrt{(n-1)n} = \sqrt{\frac{n}{n-1}} $$

wangjiezhe
  • 479