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Prove, using Jensen's formula if $f(x) = x^\alpha, (\alpha > 1)$ is convex.$$\left(\sum_{i=1}^n x_i \right)^\alpha \leq n^{\alpha-1} \left(\sum_{i=1}^n x_i^\alpha \right)$$

Any tips on how to get this done? I can't really see much of a connection between Jensen's inequality and this one. It may be silly, but I'm really stuck here.

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By Jensen's Inequality, if $f(x)$ is convex, $$\frac{1}{n}\sum_{i=1}^n f(x_i) \geq f\left(\frac{1}{n}\sum_{i=1}^n x_i\right) \implies\frac{1}{n}\sum_{i=1}^n x_i^\alpha\geq \left(\frac{1}{n}\cdot \sum_{i=1}^n x_i\right)^\alpha $$ So multiplying both sides by $n^\alpha$ gives you the result.

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  • $\begingroup$ is there a reason why you decided to go with 1/n instead of $n^{\alpha-1}$? $\endgroup$ – gurski May 6 '17 at 13:13
  • $\begingroup$ I just followed the definition of Jensen's Inequality (the left hand part of $\implies$), and plugged in $f(x) = x^\alpha$. $\endgroup$ – Lazy Lee May 6 '17 at 14:02

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