## Using Area to Prove the Arithmetic-Geometric Mean Inequality

If we have real numbers $a$ and $b$, we call  $\frac{a + b}{2}$ the arithmetic mean (AM) and $\sqrt{ab}$ the geometric mean (GM) of $a$ and $b$.  In this post, we are going to examine the relationship of these two means.

To start off, let’s have a few examples.  If $a = 3$ and $b = 12$, then $GM = 6$ and $AM = 7.5$;  if $a = 4$ and $b=16$, then $GM = 8$ and $AM = 10$; if $a = 3$ and $b = 27$, then $GM = 9$ and $AM = 15$. What do you observe? Try a few example and see if your observations hold.

From a few examples above, and from your trials, you have probably observed that $GM \leq AM$ which means that $\sqrt{ab} \leq \frac{a + b}{2}$ for some positive real numbers $a$ and $b$.  This is actually true for all positive real numbers $a$ and $b$.  In the following discussion, we are going to use the concept of area to prove that the statement is true.

To begin the proof, we construct a square with side length $a + b$ made up of four rectangles and a square at the center (technically, a square is also a rectangle). Clearly, the area of each of the four rectangles is $ab$, and the square at the center has are $(a - b)^2$ (Can you see why?).  If we remove the square at the center, the remaining area is represented by the equation $(a + b)^2 - (a - b)^2=4ab$. Note that $4ab$ is the total area of the four rectangles. » Read more