Deriving AM-GM in a way that reflects my intuition.

Question

One of my core intuitions with inequalities (especially the olympiad kind in which this is very explicit, but this has more profound influences on how I understand the relationship between inequalities) is that in a homogenous inequality, on average, terms that are more mixed ($x^2yz$) are smaller than their less-mixed counterparts ($x^4$).

With this bit of intuition, many inequalities seem trivially true, like am-gm:

$$x^3+y^3+z^3\geq xyz + xyz +xyz$$

Or this one:

$$x^2y + y^2z + z^2x\geq xyz+xyz+xyz$$

And a generalisation of part of this idea: Murihead's Inequality.

I want to find a proof (by which I mean not particularly formal derivation) of AM-GM, that adequately captures 'mixed terms being smaller on average' as being the reason for the inequality. I think this would satisfy some innate and inexplicable urge.

There may not exist a satisfying answer to this question (though I would certainly no one if I saw it), but hopefully, I have communicated the kind of thing I'm looking for.

Answer 1

For cyclic inequalities the Muirhead's idea does not work.

For example $$(4,3,0)\succ(4,2,1),$$ but the inequality $$\sum_{cyc}x^4y^3\geq\sum_{cyc}x^4y^2z$$ is wrong for positives $x$, $y$ and $z$ (take $y<z$ and $x\rightarrow+\infty$).

Actually.

If for non-negatives $\alpha$, $\beta$, $\gamma$, $\delta$ and $\epsilon$ $$(\alpha,\beta,0)\succ(\gamma,\delta,\epsilon)$$ and the inequality $$\sum_{cyc}x^{\alpha}y^{\beta}\geq\sum_{cyc}x^{\gamma}y^{\delta}z^{\epsilon}$$ is true for any positives $x$, $y$ and $z$, so we can prove it by AM-GM.

This fact was proved by Vasc more than $40$ years ago.

Answer 2

Here is one approach to thinking about these inequalities that I found.

It's not exactly a proof, but built off a more explicable intuition.

Consider, for example, the inequality:

$$x^3+y^3+z^3\geq xyz + xyz +xyz$$

On each side, we have the sum of the volumes of three rectangular prisms. Since the inequality is homogenous, we can interpret this as meaning that both sides have the same total surface area.

But, of any rectangular prism, a cube has the highest ratio of volume : surface area, so the left-hand-side is maximal.

This generalises easily to inequalities like:

$$x^2y+y^2z+z^2x\geq xyz + xyz +xyz$$

The less mixed terms on the left-hand-side have a higher volume : surface area ratio. It also genralises to higher dimensions.

Answer 3

I think there is a proof satisfying your intuition. I will talk just about the common AM-GM for three numbers $x^3+y^3+z^3\geq 3xyz$ but you can straightforwardly generalize the idea for more variables.

Very intuitive rearrangement inequality can be generalized for three ordered sequences of nonnegative numbers. For $a_1 \geq a_2 \geq a_3\geq 0$ and $b_1 \geq b_2 \geq b_3\geq 0$ and $c_1 \geq c_2 \geq c_3\geq 0$ the generalized rearrangement inequality states that $$a_1b_1c_1 + a_2b_2c_2 + a_3b_3c_3 \geq a_{\sigma(1)}b_{\tau(1)}c_1 + a_{\sigma(2)}b_{\tau(2)}c_2 + a_{\sigma(3)}b_{\tau(3)}c_3$$ for any permatutions $\sigma, \tau:\{1,2,3\}\rightarrow\{1,2,3\}$.

Since the AM-GM inequality is symmetric, we may assume that $x\geq y\geq z$. Then AM-GM is an immediate consequence of the generalized rearrangement inequality for three ordered sequences $x\geq y\geq z\geq 0$, $x\geq y\geq z\geq 0$, $x\geq y\geq z\geq 0$ since on the LHS of AM-GM we have the greatest possible sum of products and on the RHS we have some "mixture" of products.

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You may wonder how to (intuitively) prove rearrangement inequality and its generalization. If you read the paragraph called Intuition on Wikipedia, it should be obvious why rearrangement inequality holds for two sequences. Nevertheless I will present slightly unusual proof with something I call adding level by level (someone told me it is actually Abel's summation but I admit I was lazy to check it).

I will prove the following inequality (all numbers are nonnegative) $$\color{red}{a_1b_1} + \color{fuchsia}{a_2b_2} + \color{maroon}{a_3b_3} \geq \color{blue}{a_1b_2} + \color{lime}{a_2b_3} + \color{green}{a_3b_1}$$ and I believe you will understand how it works for any other permutation. I start with $b_3$-level (the first line which is actually equality), the second line is $b_2$-level and the third is $b_1$-level. \begin{align} \color{red}{a_1b_3}\phantom{-x_y)} + \color{fuchsia}{a_2b_3}\phantom{-x_y)} + \color{maroon}{a_3b_3} &\geq \color{blue}{a_1b_3}\phantom{-x_y)} +\color{lime}{a_2b_3} +\color{green}{a_3b_3} \\ \color{red}{a_1(b_2-b_3)} + \color{fuchsia}{a_2(b_2-b_3)} \phantom{+x_yx_y} &\geq \color{blue}{a_1(b_2-b_3)}\phantom{+x_yx_y} +\color{green}{a_3(b_2-b_3)} \\ \color{red}{a_1(b_1-b_2)}\phantom{+ a_2(b_2-b_3) +x_yx_y} &\geq \phantom{a_1(b_2-b_3) + x_yx_y+} \color{green}{a_3(b_1-b_2)} \\ \end{align} Just observe that we get the desired inequality after summing all these three "levels".

The idea for three sequences is exactly the same! Assume we want to prove this inequality $$\color{red}{a_1b_1c_1} + \color{fuchsia}{a_2b_2c_2} + \color{maroon}{a_3b_3c_3} \geq \color{blue}{a_1b_2c_3} + \color{lime}{a_2b_3c_1} + \color{green}{a_3b_1c_2}.$$ Again observe that it is just a sum of these three inequalities that follow from the rearrangement inequality for two sequences: \begin{align} \color{red}{a_1b_1c_3}\phantom{-x_y)} + \color{fuchsia}{a_2b_2c_3}\phantom{-x_y)} + \color{maroon}{a_3b_3c_3} &\geq \color{blue}{a_1b_2c_3}\phantom{-x_y)} +\color{lime}{a_2b_3c_3}\phantom{x} +\color{green}{a_3b_1c_3} \\ \color{red}{a_1b_1(c_2-c_3)} + \color{fuchsia}{a_2b_2(c_2-c_3)} \phantom{+x_yx_yx_y} &\geq \phantom{x_yx_y++} \color{lime}{a_2b_3(c_2-c_3)} + \color{green}{a_3b_1(c_2-c_3)} \\ \color{red}{a_1b_1(c_1-c_2)}\phantom{+ a_2b_2(c_2-c_3) +x_yx_yx_y} &\geq \phantom{x_yx_y++} \color{lime}{a_2b_3(c_1-c_2)}\\ \end{align}