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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.

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  • $\begingroup$ 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 $\endgroup$ – b00n heT Sep 25 '19 at 5:23
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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 $v_i \gt 0$.

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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.

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