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I want to show the following inequality, which seems like it should have an elementary proof (or even a well-known name).

Suppose $p, q$ are discrete probability distributions. And suppose that $p_i \leq x q_i$ for some $x > 1$. Then I want to show that for $r > 1$ we have the inequality: $$ \sum p_i^r \leq x^{r-1} \sum q_i^r $$

If we take away the requirement that $p$ is a probability distribution, then of course we have instead $\sum p_i^r \leq x^r \sum q_i^r$.

EDIT: This is not true. The best inequality one can show is the obvious one $\sum p_i^r \leq x^r \sum q_i^r$.

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    $\begingroup$ Hmm, is this a counterexample to the inequality, or am I missing something? The support size is $n+1$. Let $$p = (1,0,\dots,0)$$ and $$q = (\frac{1}{2},\frac{1}{2n},\dots,\frac{1}{2n}).$$ Here $x = 2$ and think of $n \to \infty$. Then $\sum p_i^r = 1$ and $\sum q_i^r = \frac{1}{2^r} + n\left(\frac{1}{2n}\right)^r$. So we get $1 \leq 2^{r-1} \cdot \frac{1}{2^r} + O(\frac{1}{n^{r-1}})$, or $$1 \leq \frac{1}{2} + O(\frac{1}{n^{r-1}}),$$ a contradiction. $\endgroup$ – usul Jul 14 '15 at 15:23
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(Comment --> answer)

The inequality unfortunately fails to hold, a counterexample is $$p = (1,0,\dots,0)$$ and $$q = \left(\frac{1}{2},\frac{1}{2n},\dots,\frac{1}{2n}\right),$$ where the support size is $n+1$ and $x=2$.

Then $\sum p_i^r = 1$, but $x^{r-1}\sum q_i^r \approx \frac{1}{2}$: \begin{align} \sum q_i^r &= \frac{1}{2^r} + n\left(\frac{1}{2n}\right)^r \\ &= \frac{1}{2^r}\left(1 + \frac{1}{n^{r-1}}\right) \\ &\to \frac{1}{2^r} \end{align} as $n \to \infty$, so $x^{r-1}\sum q_i^r \to \frac{2^{r-1}}{2^r} = \frac{1}{2}$.

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