# Prove that $(\sum_{i=1}^m (x_{i})^{p})^{q}\ge(\sum_{i=1}^m(x_{i})^{q})^{p}$ for all $x_i\in\mathbb R^+$ and $p,q\in\mathbb R^+$ such that $p \le q$

I'm studying the norm $$ℓ_∞$$. It leads me to prove the following inequality

$$\left( \sum_{i=1}^m (x_{i})^{p} \right)^{q} \ge \left( \sum_{i=1}^m (x_{i})^{q} \right)^{p}, \quad (x_1, \ldots,x_m) \in {(\mathbb R^+)}^m, \quad p,q \in \mathbb R^+, \quad p \le q$$

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

Lemma: $$x + y \ge (x^r + y^r)^{1/r}, \quad (x,y) \in {(\mathbb R^+)}^2, \quad 1 \le r \in \mathbb R$$

It is easy to check that our desired inequality is equivalent to

$$\sum_{i=1}^m x_{i} \ge \left( \sum_{i=1}^m (x_{i})^{r} \right)^{1/r}, \quad (x_1, \ldots,x_m) \in {(\mathbb R^+)}^m, \quad 1 \le r \in \mathbb R$$

We proceed to prove this one by induction on $$m$$. Clearly, it holds for $$m=1$$. Assume that it holds for some $$m \ge 1$$. Then we have

\begin{aligned} \sum_{i=1}^{m+1} x_{i} &= \left (\sum_{i=1}^{m} x_{i} \right ) + x_{m+1} \\ &\ge \left( \sum_{i=1}^m (x_{i})^{r} \right)^{1/r} + x_{m+1} \quad \text{by inductive hypothesis} \\ &\ge \left (\sum_{i=1}^m (x_{i})^{r} + (x_{m+1})^r \right)^{1/r} \quad \text{by Lemma} \\ &= \left( \sum_{i=1}^{m+1} (x_{i})^{r} \right)^{1/r} \end{aligned}

This completes the proof.

• It's correct of course. – Jakobian Aug 8 at 12:48
• Thank you so much for your verification @Jakobian :) – Abstract Analysis Aug 8 at 13:15

Indeed, let $$\frac{q}{p}=r$$, $$x_i^{p}=a_i$$ and $$f(x)=x^{r}.$$
Thus, $$f$$ is a convex function and we need to prove that $$f\left(\sum_{i=1}^na_i\right)\geq\sum_{i=1}^nf\left(a_i\right)$$ or $$f\left(\sum_{i=1}^na_i\right)+f(0)+...+f(0)\geq\sum_{i=1}^nf\left(a_i\right),$$ which is true by Karamata because $$(a_1+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n),$$ where $$a_1\geq a_2\geq...\geq a_n.$$