Give a proof for each statement.

If a group of 9 kids have won a total of 100 trophies, then at least one of the 9 kids has won at least 12 trophies.

If a person buys at least 400 cups of coffee in a year, then there is at least one day in which the person has bought at least two cups of coffee.

The average of three real numbers is greater than or equal to at least one of the numbers.

If a group of 9 kids have won a total of 100 trophies, then at least one of the 9 kids has won at least 12 trophies. If a person buys at least 400 cups of coffee in a year, then there is at least one day in which the person has bought at least two cups of coffee. The average of three real numbers is greater than or equal to at least one of the numbers.

Explanation:

1)

Suppose that each kid has less than 12 trophies

Total trophies = 100

Maximum trophies won by one kid = 11

total kids = 9

total number of trophies = 9 * 11 = 99 which contradicts the fact the total number of trophies are 100

2)

Suppose that person has less than 2 cups of coffee a day

Total cups of coffee = 400

he has bought at least one cup of coffee each day

which means

total number of cups of coffee = 1 * 366 = 366 which contradicts the fact the person buys at least 400 cups of coffee in a year

3)

Average of three number = (a + b + c) / 3

suppose that there are real numbers a, b, and c such that all three numbers are less than the average of the three numbers.

Let m be the average (a+b+c) / 3 = m. Then our assumption states that (a < m) and (b < m) and (c < m). By adding all the inequalities we get a + b + c < 3m. But m is defined to be (a+b+c) / 3, so a + b + c = 3m. But now we have that 3m = a + b + c < 3m. So 3m < 3m which is an obvious contradiction. Thus our claim is true