(assuming $n$ is finite)

This seems like an easy proof, but how could one write it down nicely?

I was thinking about proving it by cases: if one $s_i$ is negative, say $s_k$ than the statement is trivially true, with $S>s_k$,

If all $s_i$ are positive, then the statement is also true because the sum is greater than the parts, but I don't know how one would write this formally? (i.e. it seems like it follows from the properties of adding positive numbers together, but I don't know how one could explain this using symbols?)

Alternatively, is there a different way to prove this beside by two cases?

  • 3
    $\begingroup$ $\displaystyle S=\frac1{n-1}\sum_{k=1}^n(S-s_k)$, so… $\endgroup$ – Saad Mar 14 at 2:33
  • $\begingroup$ The two cases proof written is an acceptable proof imo. $\endgroup$ – Alberto Takase Mar 14 at 2:34

You can prove it quickly by contradiction:

  • Assume $S \leq s_i$ for all $i=1, \ldots , n$, but $S>0$ and $\boxed{n \geq 2}$:

Then you have $$\Rightarrow 0 < n S \leq \sum_{i=1}^n s_i = S \stackrel{S>0}{\Longrightarrow }\boxed{n \leq 1}$$

This is a contradiction.

So, there must be some $i$ with $S > s_i$.

  • $\begingroup$ This was nice. Thanks $\endgroup$ – user106860 Mar 14 at 5:48
  • $\begingroup$ @user106860 You are welcome :-) $\endgroup$ – trancelocation Mar 14 at 6:32

Note that this only holds for $n > 1$ (otherwise $S = s_1$ but is not greater).

Let $s^*$ be the minimum of the $s_i$. If $s^* \leq 0$ then we are done because $S > 0 \geq s^*$. So suppose $s^* > 0$. Let's try contradiction.

Suppose $S \leq s_i$ for all $i \in \{1, 2, \ldots, n\}$, with $n > 1$. Then in particular, $S \leq s^{*}$. We also have that

$$ S = \sum_{k=1}^{n} s_i \geq \sum_{k=1}^{n} s^{*} = ns^{*} $$

This gives us $s^{*} \geq S \geq ns^{*} \Longrightarrow s^{*} \geq ns^{*}$. Since $s^* > 0$, when we divide both sides of the inequality by $s^*$ we get $n \leq 1$, which contradicts our assumption that $n > 1$.


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