Suppose $0 \leq x_1, \dots, x_n \leq M$ where $M > 0$. If $$M-\frac{1}{n}\sum_{i=1}^n x_i < \epsilon$$
Then is there a function $k(\epsilon)$ such that $M-x_i \leq k(\epsilon)$ for all $i=1, \dots, n$ such that $\lim_{\epsilon \to 0}k(\epsilon ) = 0$?
In other words, if the average mean can be made close to the maximum, then the data are close to the maximum.
I tried triangle inequality but I must be missing some basic inequality.
$M-\frac{1}{n}\sum\limits_{i=1}^n x_i = \frac{1}{n}\sum\limits_{i=1}^n \left(M-x_i \right)$,
so $M-\frac{1}{n}\sum\limits_{i=1}^n x_i < \epsilon \implies \sum\limits_{i=1}^n \left(M-x_i \right) <n\epsilon$.
Since $M-x_i \ge 0$ for all $i$,
and the sum of non-negative reals is at least as large as the largest summand,
this means you can say $M-x_i <n\epsilon$ for all $i$.
For fixed $n$, you have $\lim\limits_{\epsilon \to 0} n\epsilon =0$.