# Bounding the sum of powers

Suppos we have $x_i\to 0$ for $i=1,\dots, q$, where $x_i>0$. I want to show that $$\sum_{i=1}^q x_i^3 = O\left(\left(\sum_{i=1}^q x_i^2\right)^{3/2}\right),$$ i.e. I wish to show that there exists a constant $M^*<\infty$ such that $$\sum_{i=1}^q x_i^3\leq M^* \left(\sum_{i=1}^q x_i^2\right)^{3/2}.$$

Which inequality can be applied in this (or even more general cases)? I first thought of Hoelder's inequality but this doesn't hold here...

What I have done: If $x_1\equiv \cdots \equiv x_q$ the assertion obviously holds.

For $q = 1$ we have $x_1^3 \leq M x_1^3$ and the assertion holds.

For $q=2$ we need to show that there exists a constant $M<\infty$ such that $$\left(\sum_{i=1}^2 x_i^3\right)^2 =x_1^6 + x_2^6 + 2x_1^3 x_2^3 \leq M \left(\sum_{i=1}^q x_i^2\right)^3 = M (x_1^6+x_2^6 + 3 x_1^4x_2^2 + 3 x_1^2x_2^4)$$ Letting $M=1$ we see that this inequality holds if, $$2x_1x_2 \leq 3x_1^2 +3x_2^2.$$ Let $x_1>x_2$ then $$2x_1x_2\leq 2x_1^2 \leq 3x_1^2+3x_2^2.$$ and the assertion holds. Similar $x_1<x_2$.

For $q>2$ I am however missing an obvious argument here: Under what conditions on $p$ and $q$ do we have $\left(\sum_{i=1}^q x_i^p\right)^q =\sum_{i=1}^q x_i^{pq} + \text{rest}$, where $\text{rest} \ge 0$?

Write $x = (x_1,\ldots,x_q)$ and adopt the notation (for ease of reading) that $$\|x\|_p = \left( \sum_{i=1}^q x_i^p \right)^{1/p}$$ for $0 < p < \infty$ and $$\|x\|_\infty = \max\{x_1,\ldots,x_q\}.$$
It is obvious (I hope) that $\|x\|_\infty \le \|x\|_p$. Then $$\|x\|_3^3 = \sum_{i = 1}^q x_i^3 \le \|x\|_\infty \sum_{i=1}^q x_i^2 = \|x\|_\infty \|x\|_2^2 \le \|x\|_2^3.$$ That is, $\|x\|_3 \le \|x\|_2$. This is the inequality you are looking for with $M^* = 1$.
This easily generalizes to exponents other than $2$ and $3$.