tightening around the mean of positive numbers Needing to convert a "proof" script in a classical mathematical proof, I encounter a difficulty with a mean tightening function.
Let $P$ a finite set of cardinal $p$ , $f$ a function of $P$ on $R^+$, having a minimum $m$ and a mean $M$ such $M>=m+1$. Let the spread $W = \sum_{x}^{\in P} |f(x)-M|$ . Feel free to replace absolute value by square function.
Now let's increase the minimum $m$ by $1$ : $m' = m + 1$ , then M becomes $M' = M+ \frac{1}{p}$ and the new spread $W' = \sum_{x}^{\in P} |f(x)-M'| = \sum_{x}^{\in P} |f(x)-M-\frac{1}{p}| $ . Here, nothing allows to sum $\frac{-1}{p}$ outside the absolute values summation.
Furthermore, this is the same than adding 1 to any other $f(x) , x \in P$ and it is still difficult to deal with the absolute value or the square function. Isolating $m$ leads to complicate expressions.
Question : How to highlight that it is only $m$ which had been incremented in the way to prove strictly that the spread decreased, $W' < W$ ? 
Subsidiary question : Is the condition $M>=m+1$ needful ?
 A: I am assuming that if there are multiple values of $x$ such that $f(x)=m$, you mean to say you are changing only one of them.  Let's say $x_0$ is such that $f(x_0)=m$ and you are changing it to $m+1$.
In computing the spread, you have $p$ different terms of the form $|f(x)-M|$.  Each of them except the one from $x=x_0$ is increasing by at most $\frac{1}{p}$, since $M$ is changing to $M+\frac{1}{p}$ but $f(x)$ is not changing.  Moreover, at least one of these terms is actually decreasing, namely any $x$ for which $f(x)$ is maximal (since then $f(x)\geq M'>M$ and so $|f(x)-M'|<|f(x)-M|$).  So there are $p-1$ terms of the form $|f(x)-M|$ for $x\neq x_0$, and the total increase in their sum is strictly less than $\frac{p-1}{p}$.  On the other hand, $|f(x_0)-M|$ is decreasing by exactly $1-\frac{1}{p}=\frac{p-1}{p}$, since $M\geq m+1$.  Thus when you include the $x_0$ term in the sum, you find that $W'<W$.
The assumption that $M\geq m+1$ is absolutely necessary here--for instance, imagine that $f(x)=m$ for all $x$.  Or, if you want $f$ to be injective, imagine the values of $f$ are all extremely close to $m$.
