Around this time, I was using the OTIS Excerpt of Evan Chen to prepare (or at least learn competitive math) for the math Olympiad. At the start of the Algebra section, I was introduced to the AM-GM inequality, of which is stated as:
Theorem 2.1 (AM-GM). For nonnegative real number $a_1,a_2,a_3,…,a_n$, we have: $$\frac{a_1 + a_2 + \cdot \cdot \cdot+a_n}{n} \geq \sqrt[n]{a_1 \cdot \cdot \cdot a_n}$$ The equality holds if and only if $a_1 = a_2 = \cdot \cdot \cdot =a_n$.
I understand the AM-GM inequality and as a whole of the RMS-AM-GM-HM chain. Until I meet the generalisation of the AM-GM in the section 2.4 of the text:
(Weighted Power Mean). Let $a_1, …, a_n$ be positive real numbers. Let $w_1, w_2, …, w_n$ be positive real numbers with $w_1 + w_2 + · · · + w_n = 1$. For any real number $r$, we define $$\mathcal{P}(r) = \begin{cases}(w_1 a_{1}^{r}+w_2 a_{2}^{r}+\cdot \cdot \cdot +w_n a_{n}^{r})^{1/r} & r \neq 0 \\ \\ a_{1}^{w_1}a_{2}^{w_2}\cdot \cdot \cdot a_{n}^{w_n}& r = 0\end{cases}$$ If $r > s$ then $\mathcal{P}(r) > \mathcal{P}(s)$. Equality occurs if and only if $a_1 = a_2 = \cdot \cdot \cdot =a_n$.
From here, I stumbled. From the very first, here is what I thought when I first read those lines:
- What is "weight" in this case is? I only know the weight in context of statistic to identify the importance of the components, but in this case, I don't actually know what it is. Looking back to the very AM-GM before, I can't find how I can convert that much generalized form, as according to the text, back to the inequality.
- Why the sum of the weight $\sum_{i = 1}^{n}w_i = 1$? What is the justification in this case?
- What happened with that two cases? Why when $r\neq 0$, we have another one?
- About the $r > s$ then $\mathcal{P}(r) > \mathcal{P}(s)$. If both are so that $r \neq 0, s \neq 0$, then won't it be $$(w_1 a_{1}^{r}+w_2 a_{2}^{r}+\cdot \cdot \cdot +w_n a_{n}^{r})^{1/r} > (w_1 a_{1}^{s}+w_2 a_{2}^{s}+\cdot \cdot \cdot +w_n a_{n}^{s})^{1/s}$$ What can we get from this?
- After all, what this serves, and what is it actually? And how to use it?
For the last bulletin, the one problem in the text is sufficient to serve as an example of the question.
Example 14 (Taiwan TST 2014). Let $a, b, c > 0$. Prove that $$3(a+b+c) \geq 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3 + b^3 + c^3}{3}}$$ Solution 14. By Power Mean with $r = 1, s =\frac{1}{3}$, and weights $ \frac{1}{9}+ \frac{8}{9} = 1$ we have the inequality $$\frac{1}{9}\sqrt[3]{\frac{a^3 + b^3 + c^3}{3}}+\frac{8}{9}\sqrt[3]{abc} \leq \frac{1}{9}\left(\frac{a^3 + b^3 + c^3}{3}\right) + \frac{8}{9}(abc).$$ Thus this is enough to prove $a^3 + b^3 + c^3 + 24abc \leq (a + b + c)^3$, which is clear.
I hope that someone will answer and provide me some advises and correct where I'm wrong, and also explain it thoroughly. If available, additional texts and resources is gladly received.
Best Answer
From the RMS-AM-GM-HM sequence to the weighted power mean there are two steps, and I think the fact that both steps are taken simultaneously is throwing you off. We'll do one at a time.
Power mean: If we disregard the geometric mean for a second, and rewrite the others a little, they fall into a clear pattern: $$\begin{array}{cc} \text{RMS}: &\left(\dfrac{x_1^2 + x_2^2 + \cdots + x_n^2}{n}\right)^{1/2}\\ \text{AM}: & \left(\dfrac{x_1^1 + x_2^1 + \cdots + x_n^1}n\right)^1\\ \text{HM}: & \left(\dfrac{x_1^{-1} + x_2^{-1} + \cdots + x_n^{-1}}n\right)^{-1} \end{array} $$ The general form here is as follows: $$ \begin{array}{cc}\text{PM:} & \left(\dfrac{x_1^r + x_2^r + \cdots + x_n^r}{n}\right)^{1/r}\end{array} $$ and this is called the power mean. RMS is $r = 2$, AM is $r = 1$ and HM is $r = -1$. This power mean is valid for all non-zero real numbers $r$ and for positive $x_i$. It turns out that it is increasing in $r$, and even strictly increasing unless all the $x_i$ are equal. Also, as it turns out, the geometric mean is the limit as $r\to 0$, so it fills that hole perfectly.
Weighted mean: If we instead do a different rewrite, another possibility of generalisation appears: $$\begin{array}{cc} \text{RMS}:& \sqrt{\frac1nx_1^2 + \frac1nx_2^2 + \cdots + \frac1nx_n^2}\\ \text{AM}: & \frac1nx_1 + \frac1nx_2 + \cdots + \frac1nx_n\\ \text{GM}: & x_1^{1/n}\cdot x_2^{1/n}\cdots x_n^{1/n}\\ \text{HM}: & \dfrac1{\frac1n\cdot\frac1{x_1} + \frac1n\cdot\frac1{x_2} + \cdots + \frac1n\cdot \frac1{x_n}} \end{array} $$ There is no mathematical reason all these $\frac1n$ have to be equal. They can be tweaked to suit other needs. Say the first value is, in one way or another, more important than the others. Then the $x_1$ should be assigned more weight. It should get more than $\frac1n$. That's the point of weighted averages. You manipulate all the so-called weights (which are originally all $\frac1n$), to get, for instance, $$ \begin{array}{cc} \text{Weighted RMS}:& \sqrt{\frac12x_1^2 + \frac1{2(n-1)}x_2^2 + \cdots + \frac1{2(n-1)}x_n^2}\\ \text{Weighted AM}: & \frac12x_1 + \frac1{2(n-1)}x_2 + \cdots + \frac1{2(n-1)}x_n\\ \text{Weighted GM}: & x_1^{1/2}\cdot x_2^{1/2(n-1)}\cdots x_n^{1/2(n-1)}\\ \text{Weighted HM}: & \dfrac1{\frac12\cdot\frac1{x_1} + \frac1{2(n-1)}\cdot\frac1{x_2} + \cdots + \frac1{2(n-1)}\cdot \frac1{x_n}} \end{array} $$ To make sure it's still a mean, which means in particular that it lies between the smallest and the largest of the $x_i$, we need to make sure that all these weights sum up to $1$, and we need to require that they are all non-negative.
And now we just do these two generalizations simultaneously to get the weighted power mean.