# Algebraic inequality proof [closed]

Let $\{p_i\}_{i=1...n}$ be a set of real numbers with $p_i \in [0,1] \; \forall i$, and $\sum_{i=1}^n p_i = 1$. I assume that there is one element in this set that is strictly larger than all the others, and wlog I call it $p_1$. I am attempting to prove or disprove the following proposition:

For all $\alpha > 1$, $p_1^{\alpha-1} > \sum_{i=1}^n p_i^{\alpha}$.

Any help appreciated.

## closed as off-topic by Namaste, C. Falcon, Arnaldo, Leucippus, Daniel W. FarlowJun 17 '17 at 2:34

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• What did you attempt prior to posting this? Please add your answer to this question in your post; otherwise it comes across as a problem statement problem with not context, no effort, etc. – Namaste Jun 16 '17 at 22:14

## 1 Answer

Note that $$p_i^{\alpha}=p_i^{\alpha-1}p_i\leq p_1^{\alpha-1}p_i$$ for all $i$, with the equality being strict for $i>1$. Therefore if $n>1$ then $$\sum_{i=1}^np_i^{\alpha}<p_1^{\alpha-1}\sum_{i=1}^np_i=p_1^{\alpha-1}$$ since $\sum_ip_i=1$.