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I'm having trouble with the following statement:

For positive variables $\{x_1,\dots,x_n\}\subset \mathbb{R_+}$ and $k\geq 1$ an integer, there exists $C=C(k)$ a numerical constant such that

\begin{align*} (\sum_n x_n)^k\leq C \sum_n x_n^k. \end{align*}

For $k=2$ this might be done by the Young inequality, what about the case $k\geq 3$?

the_candyman
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Leoncino
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    $f(y) = y^k$ is convex for $k\geq 1$. So, by Jensen's inequality, $\left(n^{-1}\sum_{i=1}^nx_n\right)^k\leq n^{-1}\sum_{i=1}^nx_n^k$. Now figure out the $n^{-1}$ on both sides to get your $C$ – lmaosome Jan 21 '21 at 15:11
  • Thanks, I realized it as well and posted it quite close to your comment, no intent of stealing. – Leoncino Jan 21 '21 at 15:12
  • the $k=2$ case in Cauchy-Schwarz and the general $k\geq 1$ is Hoelder's Inequality – user8675309 Jan 21 '21 at 22:25

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I was able to solve this question using Jensen's inequation:

The above setting is equivalent to \begin{align*} n^k\frac{1}{n^k}(\sum_n x_n)^k=n^k(\frac{1}{n}\sum_n x_n)^k\leq n^k\frac{1}{n}\sum_n x_n^k \end{align*} where the inequation comes from Jensen's inequality and $n^{k-1}$ is our $C$.

Leoncino
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