Relationship Between A.M. and G.M.

8.5 Relationship Between A.M. and G.M.

For any two positive real numbers $a$ and $b$, we have two important measures: the arithmetic mean (A.M.) and the geometric mean (G.M.). The question naturally arises — how are these two means related? Is one always larger than the other, or can they be equal? The answer reveals a fundamental inequality that appears throughout mathematics.

Let $A$ denote the arithmetic mean and $G$ denote the geometric mean of $a$ and $b$. Then

Both $a$ and $b$ are taken to be positive real numbers, so the square root is well-defined and both means are positive.

The Inequality $A \geq G$

Consider the difference between the arithmetic mean and the geometric mean:

Combine the terms over a common denominator:

The numerator is a perfect square. Recall the identity $(\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab}$. Therefore

Now, the square of any real number is always greater than or equal to zero. Since $\sqrt{a}$ and $\sqrt{b}$ are real numbers, $(\sqrt{a} - \sqrt{b})^2 \geq 0$. Consequently

This gives us the fundamental relationship:

with equality holding if and only if $\sqrt{a} - \sqrt{b} = 0$, which means $a = b$.

Worked Example: Finding Numbers from A.M. and G.M.

Example 13. The arithmetic mean of two positive numbers $a$ and $b$ is 10, and their geometric mean is 8. Find the numbers.

From the given information:

We now have the sum and product of two numbers. To find the individual numbers, we use the identity relating the square of the difference to the square of the sum and the product:

Substitute the known values:

Therefore

We now solve the system: …