Suppose $\boldsymbol{x}$ is an $n$-dimensional vector and all elements are nonnegative with the condition that $\sum_{i=1}^nx_i= a$. I am wondering how it is possible to find an upper bound on $\sum_{i=1}^n x_i^p$ where $0<p<1$. For special case where $p=1/2$, we can use Cauchy-Schwarz inequality ($\sum_{i=1}^n x_i^{1/2} \le \sqrt{\sum_{i=1}^n x_i} $), but for a general $0<p<1$, I do not know how we can derive an upper bound that depends on $a$.
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I guess it is different because I do not want to use Holder inequality where we have $1/p+1/q=1$. I want to find an upper bound based on $a$ which is the summation of all elements. – Amin Jul 18 '24 at 17:11
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1I don't think the referred question or answer requires $1/p+1/q=1$, just $0<p<q$... – Momo Jul 18 '24 at 17:18
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2I agree with @Momo; Jensens inequality is the way to go. – ViktorStein Jul 18 '24 at 17:21
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In Relations between p norms it is shown that $\Vert x \Vert_p \le n^{1/p-1} \Vert x \Vert_1$ for $0 < p < 1$. Isn't that exactly what you are looking for? – Martin R Jul 18 '24 at 17:48