Extending Cauchy-Schwarz to any $p \in (1,\infty)$

Question

Given $n$ non-negative real number $\left\lbrace a_i\right\rbrace_{i=1}^n$, we know that using Cauchy-Schwarz, one can prove that $$ \sum\limits_{i=1}^n \sqrt{a_i} \leq \sqrt{n} \cdot \sqrt{\sum\limits_{i=1}^n a_i} $$

Is it true that the same would apply to any $p \in (1,\infty)$, i.e. $$ \sum\limits_{i=1}^n \sqrt[p]{a_i} \leq \sqrt[p]{n} \cdot \sqrt[p]{\sum\limits_{i=1}^n a_i} \ ? $$

If not, can we multiply the right term by some constant depending on $p$ which will result in having the same form of the inequality?

Please advise.

Answer 1

You are probably looking for https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure.

Answer 2

By Holder's inequality, for $\frac{1}{p}+\frac{1}{q}=1$, $1\le p,q$, $$\sum_{i=1}^n\sqrt[p]{a_i}\le\|1\|_q\|(\sqrt[p]{a_i})\|_p=\sqrt[q]{n}\sqrt[p]{\sum_{i=1}^na_i}$$ Note that the root power of $n$ is not $p$ (except when $p=2$).

Answer 3

It's wrong.

Try $n=p=3$ and $a_1=a_2=a_3=1.$

We need to prove that $$3\leq\sqrt[3]3\cdot\sqrt[3]3,$$ which is impossible.

By the way, by Holder $$n^{p-1}\sum_{k=1}^na_k\geq\left(\sum_{k=1}^n\left(a_k\right)^{\frac{1}{p}}\right)^p,$$ which gives $$n^{1-\frac{1}{p}}\left(\sum_{k=1}^na_k\right)^{\frac{1}{p}}\geq\sum_{k=1}^n\left(a_k\right)^{\frac{1}{p}} .$$