Q:

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

Accepted Solution

A:
Answer:The given statement is true.         Step-by-step explanation:We are given the following information in the question:The average of three real numbers is greater than or equal to at least one of the numbers.We can prove this statement.Let a, b and c be the three real numbers.Then, the average of these three real numbers is given by [tex]\text{Average} = \displaystyle\frac{\text{Sum of all observation}}{\text{Total number of observation}}\\\\A = \frac{a+b+c}{3}\\\\3A = a + b + c[/tex]Let a be the smallest of the three natural number.Then, we can write,[tex]a + b+c \geq a+a+a\\3A \geq 3a\\A\geq a[/tex]Let a be the largest number.[tex]a + b+c \leq a+a+a\\3A \leq 3a\\A\leq a[/tex]Thus, looking at both the inequalities, we can say that the average of three real numbers is greater than or less than equal to one of the numbers.