I haven’t blogged here for quite a while, I hope this blog has not been abandoned. Anyway this is the first post since I have been in Australia. This is actually a note I prepared when I was in undergraduate year two, about the power mean inequality (I’ve forgotten about this note and found it in my computer recently). More precisely these are all the exercises in a chapter of a book whose name I’ve forgotten. The stuff is quite elementary but perhaps some of you are interested. If I have to make this note now, I think I will organize it in a different way, since it seems that this is organized quite loosely. But since I do not have plenty of time, I just do little formatting and put it here.

The quantities that provide the upper bound in AM-GM inequality are special cases of the general means

where is a vector of positive weights with total mass of and is a vector of positive real numbers. Unless otherwise stated, we will hereafter assume this condition for **x** and **p**. As we will see, there is a reasonable definition for the formula (1) at and at so that the map is continuous at .

Proposition 1 (The Geometric Mean as a Limit)The geometric mean can be regarded as the limit of the general mean,

This quantity is defined to be , so that the map is continuous at .

*Proof:* Consider

Taking the limit into and then applying exponential, we can get the result.

Proposition 2(Power Mean Bound for the Geometric Mean)

*Proof:*

We aim at the inequality (Siegel’s doubling relation)

By Cauchy’s inequality and the splitting trick , we have

Taking the -th root on both sides, we get (3). We deduce that for all ,

Taking the limit ,

Siegel’s doubling relation (3) and the following plot of the two-term power mean for

suggests the following inequality:

Theorem 3 (Power Mean Inequality)The mapping is a nondecreasing function on , i.e.

The equality holds if and only if

*Proof:* For , by Jensen’s inequality for the map with ,

With this inequality and the substitutions , gives

so taking -th root gives the desired inequality in this case. Moreover, the strict convexity of for ensures that the equality holds if and only if .

For , we have , then by the above,

For . As , by (2),

As , together with (2) gives the results. For the remaining cases and , they have been covered by (2) or (4) already.

Finally, as the figure suggests, we define

It is easy to show that We also have the continuity relations

For and all ,

As , as , so for all ,

So

Therefore

By for and by the first continuity relation, the second continuity relation follows.

Proposition 4 (Harmonic Means and Recognizable Sums)Suppose are positive and denotes their sum,

*Proof:* By AM-HM inequality,

Exercise 1 (Polya’s Minimax Characterization)Suppose you have to guess for an unknown x in the interval , it is possible to minimize the largest possible relative error of your guess. That is, forthere is a such that

*Proof:* Firstly, for a given , we can differentiate with respect to to see that only when , which is when the relative error is zero. Therefore for to attains its maximum, must be equal to or . Therefore

Then

Let , which is the harmonic mean of and with equal weights. It is easy to see that is increasing for and is decreasing for . Therefore attains its minimum at .

Proposition 5 (The Geometric Mean as a Minimum)The geometric mean has the representationwhere is the region of defined by

*Proof:* Assume for all , by AM-GM inequality,

The AM-GM inequality also tells us that the equality holds if and only if then and . So choosing , we obtain the result.

This result can be used to prove that geometric mean is superadditive, i.e.

Proposition 6 (More on the Method of Halves)

*Proof:* By ,

So

But

So we get the result.

Substitute in the identity and using the formula , we have the identity

Proposition 7 (A Niven-Zuckerman Lemma for -th Powers)Consider a sequence of -tuples of nonnegative real numbers , . Suppose there is a for whichand suppose for some ,

Then for all ,

*Proof:* By the power mean inequality, as , for all ,

Then

So

By Proposition 3 we know that this equality holds if and only if

But as , so , i.e.