Question

If both x and y are positive distinct real numbers, show that the geometric mean between x and y is less than their arithmetic mean?

Answer 1

To prove that the geometric mean of two distinct positive real numbers is less than their arithmetic mean, one must square the inequality comparing the means and then rearrange and simplify the terms to show that the inequality holds true.

Step-by-step explanation:

To show that the geometric mean between two distinct positive real numbers is less than their arithmetic mean, we can use an inequality and algebraic manipulation.

Let x and y be two distinct positive real numbers. The geometric mean is given by the square root of their product √(xy), while the arithmetic mean is given by the average of the two numbers ½(x + y).

We want to prove:

• √(xy) < ½(x + y)

Squaring both sides of the inequality we have:

• xy < ¼(x + y)²

Expanding the right side:

• xy < ¼(x² + 2xy + y²)

Rearranging terms, we get:

• 0 < ¼(x² - 2xy + y²)

This simplifies to:

• 0 < ¼(x - y)²

Since the square of any real number is non-negative and x and y are distinct, (x - y)² > 0, which proves the initial inequality.