# Soo, fix 0 &#x2264;<!-- ≤ --> &#x03F5;<!-- ϵ --> &#x2264;<!-- ≤ --> 1 . Given &

Soo, fix $0\le ϵ\le 1$. Given $\lambda \ge 1,{x}_{i}\ge 0$, I know that $\sum _{i=1}^{n}{x}_{i}\ge \lambda n$. I also know that $\sum _{i=1}^{n}{x}_{i}^{2}\le {\lambda }^{2}n+ϵ$. I am trying to prove a multiplicative error on each ${x}_{i}$, mainly something along the lines of
$|{x}_{i}-\lambda |\le f\left(ϵ\right)\lambda$
Where $f\left(ϵ\right)$ is some function of $ϵ$, say $f\left(ϵ\right)=2ϵ$. Is there any inequality that would bound that distance?
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Zachery Conway
Here is the solution which I literally thought of two minutes after typing the question. Let ${x}_{i}=\lambda +{\delta }_{i}$ for some ${\delta }_{i}$. Then we have that
$\sum _{i}\left({x}_{i}^{2}\right)=\sum _{i}\left(\lambda +{\delta }_{i}{\right)}^{2}={\lambda }^{2}n+2\lambda \sum _{i}{\delta }_{i}+\sum _{i}{\delta }_{i}^{2}<{\lambda }^{2}n+ϵ$
But we have that
$\sum _{i}{x}_{i}=\sum _{i}\lambda +{\delta }_{i}=\lambda n+\sum _{i}{\delta }_{i}>\lambda n$
Which implies
$\sum _{i}{\delta }_{i}>0$
Combining with first inequality
${\lambda }^{2}n+0+\sum _{i}{\delta }_{i}^{2}<{\lambda }^{2}n+2\lambda \sum _{i}{\delta }_{i}+\sum _{i}{\delta }_{i}^{2}<{\lambda }^{2}n+ϵ$
Or simply
$\sum _{i}{\delta }_{i}^{2}<ϵ$
Or
$|{\delta }_{i}|\le \sqrt{ϵ}\le \sqrt{ϵ}\lambda$