A Proof of the Arithmetic Mean Geometric Mean Inequality

Have you ever wondered what is the relationship between the arithmetic mean and the geometric mean of two numbers? Or you have probably heard some theorems about their relationship. In this post, we are going to see that their relationship is really very easy to imagine if represented geometrically.

The arithmetic mean of two numbers $latex a$ and $latex b$ is $latex \frac{a+b}{2}$, while their geometric mean is $latex \sqrt{ab}$. Now, we represent them using the figure below.

7 December 2013, Created with GeoGebra

Consider the semi-circle with diameter $latex \overline{AB}$ and radius $latex \overline{EF}$. We construct $latex \overline{CD} \perp \overline{AB}$ anywhere such that $latex C$ is on the circle and $latex D$ is on the diameter.If we let the length of $latex \overline{AB} = a$ and $latex \overline{BD} = b$, then the radius $latex \overline{EF} = (a+b)/2$.

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Arithmetic Mean Proof

In this post, we show that if $latex a$ and $latex b$ are real numbers and $latex a$ is less than $latex b$, then their arithmetic mean is greater than $latex a$. That is, $latex a < b$ implies

$latex a < \displaystyle \frac{a + b}{2}$.

Proof 1: Direct Proof

We know that $latex a < b$.

Dividing both sides by 2, we have

$latex \displaystyle \frac{a}{2} < \displaystyle \frac{b}{2}$.

Adding $latex \frac{a}{2}$ to both sides results to  Continue reading