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Let $a_1,...,a_n$ be positive real numbers and let $0 < p < 1$. Then $$(a_1 + \cdots + a_n)^p \leq a_1^p + \cdots + a_n^p. $$ Now take $ 1 \leq p < \infty$. Can we get a similar inequality, like $$ (a_1 + \cdots + a_n)^p \leq C_p(a_1^p + \cdots + a_n^p),$$ where $C_p > 0$ is a constant only depending on $p$? If not, what is the better approach we can get?

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    $\begingroup$ Take $a_1=\cdots=a_n=1$… $\endgroup$ – Saad Jun 9 '18 at 9:31
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    $\begingroup$ @AlexFrancisco $C_p=n^{p-1}$ works and is tight. Proof could be by using convexity or mean inequalities like Power Means. $\endgroup$ – Macavity Jun 9 '18 at 9:32
  • $\begingroup$ @Macavity $C_p$ should be independent from $n$. $\endgroup$ – Saad Jun 9 '18 at 9:33
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    $\begingroup$ @AlexFrancisco If that’s intended there is no $C_p$. $\endgroup$ – Macavity Jun 9 '18 at 9:34
  • $\begingroup$ Once you have the educated guess that $C_p=n^{p-1}$, Jensen's Inequality fairly rapidly proves that this is indeed the case by convexity of $x^p$. $\endgroup$ – Robert Wolfe Jun 9 '18 at 14:20
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Let $f(x)=x^p$, $0<p<1$ and $a_1\geq a_2\geq...\geq a_n$.

Thus, $f$ is a concave function and $(a_1+a_2+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n)$.

Thus, by Karamata we obtain: $$f(a_1+a_2+...+a_n,0,...,0)\leq f(a_1)+f(a_2)+...f(a_n)$$ or $$(a_1+a_2+...+a_n)^p\leq a_1^p+a_2^p+...+a_n^p$$ and we are done!

Also, we see that we got a best estimation: $C_p=1$.

For $p\geq1$ and $a_1=a_2=...=a_n=1$ we get $C_p=n^{p-1}.$

We'll prove that $$(a_1+a_2+...+a_n)^p\leq n^{p-1}(a_1^p+a_2^p+...+a_n^p)$$ or $$\left(\frac{a_1+a_2+...+a_n}{n}\right)^p\leq\frac{a_1^p+a_2^p+...+a_n^p}{n},$$ which is Jensen for $f(x)=x^p$.

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    $\begingroup$ If you take $a_1 = \cdots = a_n = 1$ you would have $ n^p \leq n$, and this is not true for $p > 1$. $\endgroup$ – Javier González Jun 9 '18 at 12:58
  • $\begingroup$ Yes, I fixed my post. Thank you! I thought that we need to prove that $(a_1+...+a_n)^p\leq a_1^p+...+a_n^p$ for $0<p<1$. $\endgroup$ – Michael Rozenberg Jun 9 '18 at 14:09

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