Why are AM-GM inequality related exercises so popular

Question

Everyday I visit this site I come across plenty AM-GM inequality related exercises.

Where I went to high school (Switzerland) we were never taught this inequality. I thus fail to see the need to talk so much about this inequality and consequently see these exercises as matematically very uninteresting: (judging on my impression only) they tend to lead nowhere and are just “mathematical juggling”.

As I currently teach mathematics at high-school, I want to better understand this inequality, in order to choose whether I shall treat this in class, or not.

So my question is: what are “useful” results or interesting topics that could lead me towards discussing this inequality in class?

I know this is opinion based, but I think it still fits the MSE policy.

Answer 1

This is the famous $\mathbf{RMS-AM-GM-HM}$ inequality. $\qquad$ Original video I think this is an interesting and fun exercise for high school students to understand the inequality.. $$\mathbf{Constructing AM}$$ We draw a semi-circle with diameter$ a + b$. Its radius is half the diameter, which is the arithmetic mean $(a + b)\over 2$. $$\mathbf{Constructing RMS}$$ The root-mean square is the hypotenuse of the following right triangle, in which one leg is the radius of the circle (the AM), so the RMS is never smaller than the AM. $$\mathbf{Constructing GM}$$ The geometric mean is the length of the perpendicular where a and b meet, which is never larger than the radius of the circle. $$\mathbf{Constructing HM}$$ This is the most involved construction. We construct a triangle with the geometric mean as one leg and the radius of the circle as the hypotenuse. If we draw an altitude to this hypotenuse, the upper length on the hypotenuse is the harmonic mean. As the HM is a leg of a triangle where the GM is the hypotenuse, the GM is never smaller than the HM.

$$\text{Putting it all together}$$ We have illustrated the pairwise inequalities:

• RMS ≥ AM
• AM ≥ GM
• GM ≥ HM

If $a ≠ b$, then the above are all strict inequalities, since the hypotenuse of a right triangle is strictly larger than a leg in the triangle. (And vice versa: if the inequalities are strict, then a ≠ b.)

If a = b, then by direct calculation we will have all quantities are equal to a (or b). Conversely, what if the inequalities are all equalities? Then it must be that the hypotenuse of each triangle is equal to a leg, which only happens if a = b when all lengths become the radius of the semi-circle.