# Inequality for $p>1$ with nonnegative real numbers

Let $$1 and $$x=(x_1,x_2,\ldots,x_N)\in\mathbb{R}^N$$, $$N\geq 2$$. Then does the following inequality holds: $$(|x_1|^{2(p-1)}+\ldots+|x_N|^{2(p-1)})^\frac{1}{2}\leq(|x_1|^p+\ldots+|x_N|^p)^\frac{p-1}{p}.$$

I can only see it holds for $$p=2$$. Can anyone please help to prove or disprove it for $$p\neq 2$$. Thanks.

Let the $$p$$-norm $$\|x\|_p$$ of $$x\in\mathbb{R}^N$$ be defined as $$\left(\sum_{n=1}^N |x_n|^p\right)^{1/p}$$. Notice then that the left hand side of the inequality is $$(\|x\|_{2(p-1)})^{p-1}$$, and the right hand side is $$(\|x\|_p)^{p-1}$$. Since we take $$p\gt 1$$, this implies that we can remove the powers and the inequality will be the same inequality as the one for $$\|x\|_{2(p-1)}$$ and $$\|x\|_p$$. Now since $$p\gt 1$$, there is a fact that the higher order p-norm is lesser than or equal to the lower order $$p$$-norm (intuition: higher order $$p$$-norms define a “stricter” sense of distance). Proof: let $$b\gt a\gt 1$$. Now, think about the expansion of the multinomial, when all values in the brackets are positive: $$(\|x\|_a)^b\ge\sum_{n=1}^N |x_n|^{a\cdot (b/a)}=(\|x\|_b)^b$$ which shows that the $$a$$-norm of $$x$$, the lower order norm, was greater than or equal to than the $$b$$-norm of $$x$$ - key here is the fact that both $$a,b\gt 1$$. This proves your inequality for all $$p\ge2$$ since $$2(p-1)\ge p$$ if $$p\ge2$$, so the $$p$$-norm is a lower order norm than the $$2(p-1)$$-norm, therefore the $$p$$-norm of $$x$$ is greater than or equal to, therefore the right hand side is greater than or equal to the left hand side.

It is false for $$p$$ in the range $$(1,2)$$ since in those ranges $$p\gt 2(p-1)$$, and therefore the norm in the LHS is of lower order, not higher, making the left hand side larger.

For example, take $$x=(0.5,0.5)\in\mathbb{R}^2$$, with $$p=1.5$$. The LHS is $$(0.5^1+0.5^1)^{1/2}=1$$, but the RHS is $$(0.5^{3/2}+0.5^{3/2})^{1/3}=0.89089871813\lt1\lt LHS$$.

In $$p$$-Norm notation, we have $$\|x\|_{2(p-1)}^{p-1}\leq \|x\|_p^{p-1}$$. This is equivalent to $$\|x\|_{2(p-1)}\leq \|x\|_p$$, which is true, because $$2(p-1)\geq p$$ for $$p\geq 2$$. This is true by the monotonicity of the $$p$$-norms.

We prove $$\|x\|_p\leq \|x\|_r$$ for $$r. W.l.o.g. suppose that $$\|x\|_p=1$$. Then $$\|x\|_p^p=1$$ and $$|x_i|\le 1$$ for all $$i$$. Since $$p>r$$, we obtain $$|x_i|^p \le |x_i|^r$$ for all $$i$$ and we get $$1=\|x\|_p^p = \sum_{i} |x_i|^p \le \sum_{i} |x_i|^r = \|x\|_r^r.$$ The proof of the monociticy is not my own proof. You can find it on several places in the internet.

• Amusing how we gave almost the same answer within two minutes of each other Jul 26, 2021 at 10:16
• Ha ha, yeah, i didn't realized, that there is another answer :D Jul 26, 2021 at 10:20
• @Jochen and Idiotic Shrike Thank you very much. If I understood correctly, the inequality in the question holds for $p\geq 2$ and it is false for $1<p<2$. Jul 26, 2021 at 10:21
• Well, it is only provably true for all $x$ if $p\ge 2$. Let me see if it is provably false for all $1\lt p\lt 2$ Jul 26, 2021 at 10:25