Proof that given n real numbers, that less than half of them are greater than twice the average

Question

I am not sure exactly as to how to write this proof.

For example, I tried inducting upon n numbers (with a base case of n=3)

But I was not sure if this would be the best idea. Any advice?

Answer 1

It is not true. Take $-10,0,1$. The average is $-3$. Both $0$ and $1$ are greater than $-6$. I suspect the numbers are supposed to be positive. Assume there is a set as specified. Let the average be $a$. What is the sum? What is the sum of the ones greater than half the average? Can you find a contradiction?

Answer 2

Well, if if you add up this numberst that are more than twice the average....

If the sum of then numbers is $SUM$ and there are $n$ numbers then the average is $\frac {SUM}n$. If $\frac n2$ or more are greater than $\frac {2*SUM}n$ then if you add them up you get they add up to more than $\frac n2*\frac {2*SUM}n = SUM$.

The only way a fewer than all the numbers can add up to more than what all the numbers add to is if some the numbers add up to less than zero.

If that's possible then... this is possible. If you assume that the numbers must but non-negative, this isn't possible.