Inequality for p-norm of probability distributions 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 $x &gt; 1$. Then I want to show that for $r &gt; 1$ 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 $\sum p_i^r \leq x^r \sum q_i^r$.
(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 &amp;= \frac{1}{2^r} + n\left(\frac{1}{2n}\right)^r  \\
  &amp;= \frac{1}{2^r}\left(1 + \frac{1}{n^{r-1}}\right)  \\
  &amp;\to \frac{1}{2^r}
\end{align}
as $n \to \infty$, so $x^{r-1}\sum q_i^r \to \frac{2^{r-1}}{2^r} = \frac{1}{2}$.