# relationship between sum of squares and sum

I have to admit I am not good at math since it's been a while since I did the last math problem. I am working on a project where there is a problem that can be summarized like this:

if $\sum_{i=1}^{n}a_i^2 = x$, can we determine a upper(and/or lower) bound for $\sum_{i=1}^{n}|a_i|$ ?

Sorry if this is too simple for you, I appreciate your help!

PS: Thank you for all the helps I got! I just realize that I need to know the upper bounds for both $\sum_{i=1}^{n}|a_i|$ and the individual $|a_i|$. So what is the finest upper bound for $|a_i|$ I can get? Thank you again.

• Yeah, it would be better if there is a smaller upper bound. thank you May 20, 2016 at 12:22
• Do you know if $|a_k|>1$ or $<1$? May 20, 2016 at 12:22
• yes, $|a_k|$ is a smaller number than 1. May 20, 2016 at 12:24
• then $x$ is a lower bound for the second sum May 20, 2016 at 12:26
• As for the individual $|a_i|$, it depends on how much you know about them. WIth no additional information you might as well use the same upper bound as for the whole sum, because one of the terms can be much bigger than all the others combined May 20, 2016 at 14:13

Both sums are norms, $\lVert. \rVert_2$ vs $\lVert . \rVert_1$ and thanks to norm equivalence for finite dimensional spaces there are factors which fulfill $$m \lVert a \rVert_2 \le \lVert a \rVert_1 \le M \lVert a \rVert_2$$ So having $\lVert a \rVert_2^2 = x$ we get $$m \sqrt{x} \le \lVert a \rVert_1 \le M \sqrt{x}$$ for $m =1$ and $M = \sqrt{n}$.

• Thank you for the clear explanation. May 20, 2016 at 12:42

You can use the AQM inequality: $$\frac{\displaystyle\sum\nolimits_{i=1}^n\lvert a_i\rvert}n \le \sqrt{\frac{\displaystyle\sum\nolimits_{i=1}^n a_i^2}n}=\sqrt{\frac xn}, \enspace\text{whence}\enspace \displaystyle\sum_{i=1}^n\lvert a_i\rvert\le\sqrt{nx\mathstrut}.$$

• AQM, meaning Cauchy-Schwarz? (Note that the only non sporadic use of AQM my search engine indicates is for Additive Quark Model.)
– Did
May 20, 2016 at 12:37
• No: Arithmetic-Quadratic Means. May 20, 2016 at 12:39
• You're welcome. Always glad to help! May 20, 2016 at 12:56
• @Did Aside from quadratic mean, many people also use the name root-mean square. May 20, 2016 at 16:34

Happily for you, it turns out that $\sqrt{\sum_{i=1}^n a_i^2 } \leq \sum_{i=1}^n |a_i| \leq \sqrt{n} \cdot \sqrt{\sum_{i=1}^n a_i^2 }$