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

7 December 2013, Created with GeoGebra

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

Now, move $CD$ by dragging $D$ in the figure above. What do you observe? What is the relationship between $CD$ and $EF$ ?

Yes, you are right. If we drag $D$ on the diameter of the semi-circle,

$CD \leq EF$ (*).

We already know that $EF = (a + b)/2$. We only need to get  $CD$.

Notice that $ADC$ and $CDB$ are similar triangles (Prove it!). This means that

$\frac{a}{CD} = \frac{CD}{b}$ which gives us $CD = \sqrt{ab}$.

Now, substituting this to *, we have

$\sqrt{ab} \leq \frac{a+b}{2}$

which is the Arithmetic Mean Geometric Mean Inequality.

So, the Arithmetic Mean Geometric Mean Inequality states that the Geometric Mean of two numbers is always less than or equal to their Arithmetic Mean. Geometrically speaking, the segment perpendicular to the diameter that you can draw from the diameter of a semicircle to a point on the circle is shorter or the same length or of the same length as the radius.