# Proving If $S= \sum_i^n s_i >0$ then $S>s_i$ for some i

(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?

• $\displaystyle S=\frac1{n-1}\sum_{k=1}^n(S-s_k)$, so… – Saad Mar 14 at 2:33
• The two cases proof written is an acceptable proof imo. – 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 $$\Rightarrow 0 < n S \leq \sum_{i=1}^n s_i = S \stackrel{S>0}{\Longrightarrow }\boxed{n \leq 1}$$

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

• This was nice. Thanks – user106860 Mar 14 at 5:48
• @user106860 You are welcome :-) – 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 $$n > 1$$.