More on averages

This post is a continuation of the post "Squares and rectangles", here I will generalize a bit what we discovered in that post and deduce the geometrical/arithmetical mean inequality, as well as prepare for the next post the many variables case.

In the last post we saw that if we want a bigger area of a rectangle with fixed perimeter we should aim for a square, more specifically, if sides "a" and "b" sum "p" then:

(a+b)/2 =p/2 =M =Mean

and that is the best side possible:

a*b < M*M = (a+b)/2*(a+b)/2 (Ineq. 1)

And if we a set = M-k and b = M+k

a*b = (M-k)*(M+k) = M ² - k ² (Equal. 2)

So for fixed p (and so M) a*b decreases as k increases and vice versa, or the farthest we are from the mean the worst we are.


Taking square roots on both side on Ineq.1 we have:

sqrt(a*b) < (a+b)/2

That is the geometrical mean is less then the arithmetical mean (unless it is equal when a=b).


We could ask the same question for three variables, that is if a+b+c = p is fixed then what is the choice of a, b, and c that gives the largest product?

I guess it is reasonable to say that a= b= c= p/3= (a+b+c)/3 should be the best choice, that is indeed true but it is not easy to prove!

I will prove that in my next post, hint we will use more (Equal. 2).