Jensens Inequalty

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    Akash Arjun

    help me solve this problem

    Saumik Karfa

    take $f(x)=\frac {x}{1-x}$

    the graph shows that the function is concave within $(1,\infty)$.

    take $\lambda_i =\frac 1n$

    Therefore, by jensen’s inequality:

    $\Rightarrow \displaystyle\sum_{i=1}^n \lambda_i f(x_i) \leq f\bigg(\displaystyle\sum_{i=1}^n \lambda_i x_i\bigg)$

    $\Rightarrow \frac 1n \cdot \displaystyle\sum_{i=1}^n \frac{x_i}{1-x_i} \leq f\bigg(\frac{\displaystyle\sum_{i=1}^n  x_i}{n}\bigg)$

    $\Rightarrow \frac 1n \cdot \displaystyle\sum_{i=1}^n \frac{x_i}{1-x_i} \leq \frac{\frac{\displaystyle\sum_{i=1}^n  x_i}{n}}{1-\frac{\displaystyle\sum_{i=1}^n  x_i}{n}}$

    $\Rightarrow \displaystyle\sum_{i=1}^n \frac{x_i}{1-x_i} \leq n \cdot \frac{\displaystyle\sum_{i=1}^n  x_i}{n-\displaystyle\sum_{i=1}^n  x_i} $

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