Answer to Question #262191 in Discrete Mathematics for abhi

Question

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

Answer 1

Let us prove that the average of three real numbers is greater than or equal to at least one of the numbers.

Let $a_1,a_2,a_3$ be arbitrary real numbers, and $b=\frac{a_1+a_2+a_3}3$ be their average value. Let us prove by contraposition. Suppose that $b<a_1$ and $b<a_2$ and $b<a_3.$ Then $3b=b+b+b<a_1+a_2+a_3,$ and hence $b<\frac{a_+a_2+a_3}3=b.$ We get the contradiction $b<b.$ This contradictions prove that the average of three real numbers is greater than or equal to at least one of the numbers.