Inequality for p-norm of probability distributions
(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 $x^{r-1}\sum q_i^r \to \frac{2^{r-1}}{2^r} = \frac{1}{2}$.