# Average, spending, and pigeonhole principle proof question

There are $$n$$ days, and each day I spend $$x_i$$ dollars on the $$i$$ day. $$i\leq{n}$$. I spend nonnegative dollars per day (which could be 0 and possibly be a non-integer). After $$n$$ days, I average how much money I spent. $$\bar{x}=\frac{x_1+x_2+...+x_n}{n}$$.

1. prove that on at least one of the days, I spent at least $$\bar{x}$$ dollars.

I used the pigeonhole principle here where the total number of dollars I spent = number of pigeons, and $$n$$ days = number of holes.

1. prove that on fewer than half of the $$n$$ days, I paid strictly more than 2$$\bar{x}$$ dollars.

I'm having trouble with this proof. I have trouble deciding how and if the PHP plays a role in this. Any help is appreciated.

• You don’t need the PHP for the first: if you had never spent that amount, then the average would be lower. The same reasoning applies for the second: if you had spent more than that amount on more than half the days, then the average would be higher – b00n heT Sep 25 '19 at 5:23

If you paid exactly $$2 \bar{x}+ v_i$$ dollars on exactly $$\frac n2$$ days, then you spent a total of $$\bar{x}n + \sum v_i$$ dollars total on those days. But your total spending is $$\bar{x}n$$ dollars so $$\sum v_i \leq 0$$. That's not possible if each $$v_i \gt 0$$.

Try a proof by contradiction. Suppose on $$\ge n/2$$ days you paid $$2 \bar{x}$$ dollars. Even in the best situation where you spent $$0$$ dollars on the other days, you can show that $$\frac{1}{n}(x_1 + \cdots + x_n) > \bar{x}$$ which is a contradiction.