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I want to understand what values can be simultaneously attained as the arithmetic (AM), geometric (GM), and harmonic (HM) means of finite sequences of positive real numbers. Precisely, for what points $(G, H) \in \mathbb R^2_{\geqslant 0}$, do there exist a $n$-tuple $a = (a_1, a_2, \ldots, a_n) \in \mathbb R^n_{\geqslant 0}$ such that $$ \begin{align*} \operatorname{AM}(a) &= 1, \\ \operatorname{GM}(a) &= G, \\ \operatorname{HM}(a) &= H. \end{align*} $$ Here the constraint on AM provides an implicit normalisation on the $n$-tuple. Note that the AM-GM-HM inequality implies that such an $(G, H)$ must lie in the lower right half of the unit square given by

$$ \quad 1 \geqslant G \geqslant H \geqslant 0.\tag{$\ast$}$$

I think that the problem is quite hard for a general $n$, so I will be content with $n = 4$. In fact, I will also be happy to see any bounds improving on $(\ast)$, either for general $n$ or for specific small values of $n$. Below I sketch the solution for $n= 2$ and $n=3$.


Case $n=2$. It turns out that, for any pair of numbers $(a_1, a_2)$, the three means satisfy a tight relation: $\operatorname{AM} \cdot \operatorname{HM} = \operatorname{GM}^2$. That is, all our points must lie on the parabola $H = G^2$.


Case $n=3$. Now the tight relationship between the means disappears; however, the situation is still rigid enough to permit an algebraic solution. It is easy to check that the three numbers are roots of the cubic equation $$ x^3 - 3 \operatorname{AM} x^2 + 3 \frac{\operatorname{GM}^3}{\operatorname{HM}} x - \operatorname{GM}^3 = 0, $$ which rearranges to $$ H x^3 - 3H x^2 + 3 G^3 x - G^3 H = 0. $$ We want this equation to have three real, nonnegative roots. It is well-known† that this is equivalent to the condition that the discriminant of the cubic is nonnegative: $$ \begin{array}{crl} & 27 (6 G^6 H^3+3 G^6 H^2 -4 G^9 H-G^6 H^4-4 G^3 H^4) &\geqslant 0 \\ \iff & 6 G^3 H^2+3 G^3 H -4 G^6- G^3 H^3-4 H^3 &\geqslant 0. \end{array} $$

Here is a plot of this region for $n=3$ (with $G$ along the $x$-axis and $H$ along the $y$-axis):

AM-GM-HM plot for n=3


Question. To restate my question:

Can you give any bounds on the AM-GM-HM region for any $n \geqslant 4$ that beats $(\ast)$?

In fact, the following conjecture looks plausible:

Conjecture. For any $n$, there exists $c = c_n > 0$ such that the following holds for any $n$-tuple of positive reals: $$ \operatorname{AM} + c \operatorname{HM} \geqslant (1+c) \operatorname{GM}. $$ That is, the region lies entirely to the left of the line $G = \frac{1}{1+c} + \frac{c}{1+c} H$.

This conjecture is inspired by an earlier post that essentially claims that $c = \frac{33}{50}$ works for $n=5$.‡ The proposed answer in that thread suggests using Lagrange multipliers. But unfortunately, it seems to me that it is rather too sparse on the details; I am not entirely sure how fruitful this approach would be.*


†See the Wikipedia article on solving a cubic equation.

‡The equivalence between the stated problem and my question is explained clearly in Zarrax's answer in that thread.

*The answer is now deleted.

[Thanks to QED for the plot.]

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My intuition insists that the lower bound should be achieved by the $n$-tuple $(t,t,\ldots,t,1-(n-1)t)$ for $t < 1$ (and perhaps the upper bound by the same tuple with $t > 1$). The corresponding plot for $n = 3$ looks like it supports that conjecture. –  Rahul Dec 20 '11 at 13:23
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Whoops, that reminds me, the last element of the tuple should be $n - (n-1)t$, not $1 - (n-1)t$. The valid range of $t$ is $0$ to $n/(n-1)$, if I haven't made another mistake. And yes, I think I used too wide a range in my plot; thankfully WolframAlpha ignored non-real values of the coordinates :) –  Rahul Dec 20 '11 at 13:28
    
@Rahul Thank you for the update; it looks good now. :) –  Srivatsan Dec 20 '11 at 13:33
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It turns out my intuition was mistaken in at least one respect: the curve for $t<1$ is actually above the curve for $t>1$. –  Rahul Dec 20 '11 at 13:43
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By the way, as much as I love math.SE and MO, this question makes me a little sad. Five years ago, if someone had asked me this question over tea, I would have thought about it for a while and, if I got it, I would have sent it off to the USAMO problem committee as a really challenging inequality. But now that I've answered it in public, I can't. In a few years, it's going to be very hard to write Olympiads. –  David Speyer Dec 21 '11 at 0:55

4 Answers 4

up vote 28 down vote accepted
+200

One way to look at this problem is to look for the extremes of the geometric mean given fixed arithmetic and harmonic means.

The following inequalities can be verified by taking $\log$s and using Jensen: $$ \frac{1}{\int\frac{1}{x}\;\mathrm{d}\mu}\le\exp\left(\int\log(x)\;\mathrm{d}\mu\right)\le\int x\;\mathrm{d}\mu\tag{1} $$ That is, $HM\le GM\le AM$.

Scaling the data scales all three means by the same factor, so let us assume that $$ \int x\;\mathrm{d}\mu=1\quad\text{(arithmetic mean }=1\text{)}\tag{2} $$ and $$ \int\frac{\mathrm{d}\mu}{x}=\frac{1}{h}\quad\text{(harmonic mean }=h\text{)}\tag{3} $$ and attempt to find the critical values for the $\log$ of the geometric mean: $$ \int\log(x)\;\mathrm{d}\mu\tag{4} $$ Taking variances of $(2)$, $(3)$, and $(4)$, we get that for all variations so that the $AM$ is stationary $$ \int\delta x\;\mathrm{d}\mu=0\tag{5} $$ and the reciprocal of the $HM$ is stationary, $$ \int\frac{\delta x}{x^2}\mathrm{d}\mu=0\tag{6} $$ we should also have that the $\log$ of the geometric mean is stationary: $$ \int\frac{\delta x}{x}\mathrm{d}\mu=0\tag{7} $$ Linearity considerations say that $(5)$, $(6)$, and $(7)$ imply that there are constants $a$ and $b$ so that $$ \frac{1}{x}=a+\frac{b}{x^2}\tag{8} $$ Multiplying $(8)$ by x and combining with $(2)$ and $(3)$ yields $$ 1=a+\frac{b}{h}\tag{9} $$ Solving $(8)$ for $x$ and $(9)$ for $b$, we get $$ x=\frac{1\pm\sqrt{1-4ha(1-a)}}{2a}\tag{10} $$ To maintain $(2)$, we need a combination of the two values of $x$: $$ \lambda\frac{1+\sqrt{1-4ha(1-a)}}{2a}+(1-\lambda)\frac{1-\sqrt{1-4ha(1-a)}}{2a}=1\tag{11} $$ Solving $(11)$ for $\lambda$, we get $$ \lambda=\frac{1}{2}\left(1+\frac{2a-1}{\sqrt{1-4ha(1-a)}}\right)\quad\text{and}\quad a=\frac{1}{2}\left(1+\frac{2\lambda-1}{\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}}\right)\tag{12} $$ Note that $(12)$ shows that $\lim\limits_{a\to0}\frac{\lambda}{a}=\lim\limits_{a\to1}\frac{1-\lambda}{1-a}=1$.

For a given harmonic mean, $h$, and split between values of $x$, $\lambda$, $(10)$ and $(12)$ yield that the stationary geometric mean is $$ \begin{align} g(h,\lambda) &=\left(\frac{1+\sqrt{1-4ha(1-a)}}{2a}\right)^\lambda\left(\frac{1-\sqrt{1-4ha(1-a)}}{2a}\right)^{1-\lambda}\\ &=\frac{\left(\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}+1\right)^\lambda\left(\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}-1\right)^{1-\lambda}}{\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}+2\lambda-1}\tag{13} \end{align} $$ For each $h\in[0,1]$, $g(h,\lambda)$ is monotonic in $\lambda$ (verified below) and $$ \lim_{\lambda\to0}g(h,\lambda)=h\qquad\text{and}\qquad\lim_{\lambda\to1}g(h,\lambda)=1\tag{14} $$ With a countinuous selection of $\lambda$, we can attain any value of $g(h,\lambda)$ between $h$ and $1$. However, When dealing with $n$ numbers, $\lambda$ can only take the values $\frac{1}{n}\dots \frac{n-1}{n}$. Thus, with a harmonic mean of $h$, the geometric mean can only vary between $g(h,\frac{1}{n})$ and $g(h,\frac{n-1}{n})$.

$\mathbf{n}$ objects with arithmetic mean $\mathbf{=1}$:

range of geometric means


Proof that for any $\mathbf{h}$, $\mathbf{g(h,\lambda)}$ is monotonic in $\mathbf{\lambda}$:

Fix $h$. Define $$ \Delta=\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}\tag{15} $$ Using $(13)$ and $(15)$, we get that $$ g(h,\lambda)=\frac{(\Delta+1)^\lambda(\Delta-1)^{1-\lambda}}{\Delta+2\lambda-1}\tag{16} $$ Equation $(15)$ also implies that $$ (1-h)\Delta^2+h(2\lambda-1)^2=1\tag{17} $$ In consideration of $(17)$, define $\eta$ and $\theta$ by $$ \begin{align} \sin(\eta)&=\sqrt{h}\\ \cos(\eta)&=\sqrt{1-h} \end{align}\tag{18} $$ and $$ \begin{align} \sin(\theta)&=\sin(\eta)\;(2\lambda-1)\\ \cos(\theta)&=\cos(\eta)\;\Delta \end{align}\tag{19} $$ where $0\le\lambda\le1$ and thus $-\eta\le\theta\le\eta$. Then using the identities $$ \begin{array}{ccccc} \cos(\eta)(\Delta+1)&=&\cos(\theta)+\cos(\eta)&=&2\cos\left(\frac{\eta+\theta}{2}\right)\cos\left(\frac{\eta-\theta}{2}\right)\\ \cos(\eta)(\Delta-1)&=&\cos(\theta)-\cos(\eta)&=&2\sin\left(\frac{\eta+\theta}{2}\right)\sin\left(\frac{\eta-\theta}{2}\right)\\ \sin(\eta)2\lambda&=&\sin(\eta)+\sin(\theta)&=&2\sin\left(\frac{\eta+\theta}{2}\right)\cos\left(\frac{\eta-\theta}{2}\right)\\ \sin(\eta)2(1-\lambda)&=&\sin(\eta)-\sin(\theta)&=&2\cos\left(\frac{\eta+\theta}{2}\right)\sin\left(\frac{\eta-\theta}{2}\right) \end{array}\tag{20} $$ we get that $$ \begin{align} &\left(\frac{1}{\sin(\eta)}\frac{(\Delta+1)^\lambda(\Delta-1)^{1-\lambda}}{\Delta+2\lambda-1}\right)^{2\sin(\eta)}\\ &\vphantom{\Huge{\dfrac{A}{A}}}=\frac{(\cos(\theta)+\cos(\eta))^{\sin(\eta)+\sin(\theta)}(\cos(\theta)-\cos(\eta))^{\sin(\eta)-\sin(\theta)}}{\left(\sin(\eta)\cos(\theta)+\cos(\eta)\sin(\theta)\right)^{2\sin(\eta)}}\\ &\vphantom{\Huge{\dfrac{A}{A}}}=\frac{(2\cos(\frac{\eta+\theta}{2})\cos(\frac{\eta-\theta}{2}))^{\sin(\eta)+\sin(\theta)}(2\sin(\frac{\eta+\theta}{2})\sin(\frac{\eta-\theta}{2}))^{\sin(\eta)-\sin(\theta)}}{\sin(\eta+\theta)^{2\sin(\eta)}}\\ &\vphantom{\Huge{\dfrac{A}{A}}}=\frac{(\sin(\eta+\theta)\sin(\eta-\theta))^{\sin(\eta)}(\cot(\frac{\eta+\theta}{2})\cot(\frac{\eta-\theta}{2}))^{\sin(\theta)}}{\sin(\eta+\theta)^{2\sin(\eta)}}\\ &\vphantom{\Huge{\dfrac{A}{A}}}=\left(\frac{\sin(\eta-\theta)}{\sin(\eta+\theta)}\right)^{\sin(\eta)}\left(\frac{\cos(\theta)+\cos(\eta)}{\cos(\theta)-\cos(\eta)}\right)^{\sin(\theta)}\tag{21} \end{align} $$ The logarithmic derivative of $(21)$ is $$ \begin{align} &\frac{\mathrm{d}}{\mathrm{d}\theta}\left(\sin(\eta)\log\left(\frac{\sin(\eta-\theta)}{\sin(\eta+\theta)}\right)+\sin(\theta)\log\left(\frac{\cos(\theta)+\cos(\eta)}{\cos(\theta)-\cos(\eta)}\right)\right)\\ &\vphantom{\Huge{A}}=\sin(\eta)(-\cot(\eta-\theta)-\cot(\eta+\theta))\\ &+\cos(\theta)\log\left(\frac{\cos(\theta)+\cos(\eta)}{\cos(\theta)-\cos(\eta)}\right)\\ &+\sin(\theta)\left(\frac{-\sin(\theta)}{\cos(\theta)+\cos(\eta)}+\frac{\sin(\theta)}{\cos(\theta)-\cos(\eta)}\right)\\ &\vphantom{\Huge{\dfrac{A}{A}}}=-2\cos(\eta)+\cos(\theta)\log\left(\frac{\cos(\theta)+\cos(\eta)}{\cos(\theta)-\cos(\eta)}\right)\tag{22} \end{align} $$ Dividing $(22)$ by $\cos(\theta)$ and setting $t=\dfrac{\cos(\eta)}{\cos(\theta)}$, yields that $(22)$ vanishes precisely when $$ -2t+\log\left(\frac{1+t}{1-t}\right)=0\tag{23} $$ However, $(23)$ vanishes only at $t=0$, but because $-\eta\le\theta\le\eta$, we have $t\ge\cos(\eta)=\sqrt{1-h}$. Therefore, $(22)$ doesn't vanish, and thus, $(21)$ is monotonic.

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I feel like I'm bringing rather heavy machinery in this problem, but here goes.

Rahul Narain's guess is correct and follows from the Equal Variable Theorem as proved here

The complete theorem is much more general, but for your question Corollary 1.9, Case 1 suffices:

Let $a_1, a_2, \ldots, a_n$ $(n \geq 3)$ be fixed non-negative numbers, let $0 \leq x_1 \leq x_2 \leq \cdots \leq x_n$ such that $$ x_1+x_2+\cdots+x_n = a_1+a_2+\cdots+a_n,$$ $$x_1^p + x_2^p +\cdots+x_n^p = a_1^p + a_2^p +\cdots+ a_n^p,$$ and let $E = x_1^q + x_2^q +\cdots+x_n^q$.

Case 1. $p \leq 0$ ($p=0$ yields $x_1x_2\cdots x_n = a_1a_2\cdots a_n > 0$).

(a) For $q \in (p,0) \cup (1, \infty)$, $E$ is maximal when $0 \lt x_1 = x_2 = \cdots = x_{n-1} \leq x_n$, and is minimal when $0 < x_1 \leq x_2 = x_3 = \cdots = x_n$.

(b) For $q \in (-\infty,p) \cup (0,1)$, $E$ is minimal when $0 \lt x_1 = x_2 = \cdots = x_{n-1} \leq x_n$, and is maximal when $0 < x_1 \leq x_2 = x_3 = \cdots = x_n$.

Now take $p=0$ for the geometric mean, $q = -1$ for the harmonic mean and use part b) from above to arrive at the claim.

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OK, I have the right proof now. Once I finish this one, I'll delete the others. I am proving Rahul's intuition, that the extreme value of the product occurs when all the $x_i$ but one are equal. More specifically, I will prove that the product is maximized when the multiple value is larger than the single value.

Lemma: Let $a$, $b$ and $c$ be positive reals; define $n = a+b+c$; let $P$ and $Q$ be positive reals with $PQ > n^2$. Then, on the curve $$\begin{matrix} P &=& ax+by+cz \\ Q &=& ax^{-1} + b y^{-1} + c z^{-1} \\ x,y,z &>& 0 , \end{matrix}$$ the maximal value of $x^a y^b z^c$ occurs at a point where two of $(x,y,z)$ are equal and the third is smaller.

Remark: If $PQ < n^2$, then the above equations have no real solutions; if $PQ=n^2$ then the only real solution is $x=y=z=P/n$. So the lemma discusses the interesting case.

Proof: First, notice that the equations above have all of their solutions within the box $[a/Q, P/a] \times [b/Q, P/b] \times [c/Q, P/q]$, and form a closed subset of that box. So the solution space is compact (closed subset of a bounded set), and the maximum must exist.

Let $(x,y,z)$ be a point where the two equations are satisfied. If we had $x=y=z$ then $PQ=n^2$, so it is not possible that all of $x$, $y$ and $z$ are equal.

The gradient of the function $f:=ax+by+cz$ is $(a,b,c)$. The gradient of $g:=a x^{-1} + b y^{-1} + c z^{-1}$ is $(-a x^{-2}, -b y^{-2}, -c z^{-2})$. Since $x$, $y$ and $z$ are not all equal, these vectors are not parallel. Thus, the two surfaces $f=P$ and $g=Q$ meet transversely, in a smooth curve. A tangent vector to that curve is $$(a,b,c) \times (-a x^{-2}, -b y^{-2}, -c z^{-2}) = {\LARGE (} bc (y^{-2} - z^{-2}), ac (z^{-2} - x^{-2}), ab(x^{-2} - y^{-2}) {\LARGE )} \quad (\ast)$$

Rather than optimizing $x^a y^b z^c$, we will optimize $h:= \log(x^a y^b z^c) = a \log x + b \log y + c \log z$. The derivative of $h$ along the vector $(\ast)$ is $$abc (x^{-1} y^{-2} - x^{-1} z^{-2} + y^{-1} z^{-2} - y^{-1} x^{-2} + z^{-1} x^{-2} - z^{-1} y^{-2} ) = \frac{abc}{x^2 y^2 z^2} (x-y)(x-z)(y-z). \quad (\ast\ast).$$

If $x$, $y$ and $z$ are all distinct, then $(\ast \ast)$ is nonzero, so moving along the curve $f=P$, $g=Q$ in one direction or the other will cause $h$ to increase. A more detailed analysis (omitted) shows that we have a local maximum when the largest number of $(x,y,z)$ occurs twice. $\square$

We now prove the result. Let $(x_1, \ldots, x_n)$ be any $n$-tuple of positive reals. We will show that there are $y$ and $z$ with $y+(n-1) z = \sum x_i$, $y^{-1} + (n-1) z^{-1} = \sum x_i^{-1}$, $y<z$ and $y z^{n-1} \geq \prod x_i$. In other words, $(y,z,z,z,\ldots, z)$ maximizes $\prod x_i$ subject to fixed values of $\sum x_i$ and $\sum x_i^{-1}$.

The proof proceeds in $n-2$ steps. At each step, we increase the number of times the maximal value occurs among $(x_1, x_2, \ldots, x_n)$ and increase $\prod x_i$, all while holding $\sum x_i$ and $\sum x_i^{-1}$ constant. So, suppose that $x_1=x_2=\cdots = x_k > x_{k+1}, \ldots, x_n$ for some $k$. If $k=n-1$ then we are done. If not, hold $x_{k+3}$ through $x_n$ fixed, while replacing $(x_1, \ldots, x_{k+2})$ by the values which maximize $\prod_{i=1}^{k+2} x_i$ while holding $\sum_{i=1}^{k+2} x_i$ and $\sum_{i=1}^{k+2} x_i^{-1}$ fixed. By the Lemma, the new maximal value will now occur $k+1$ times. After finitely many steps of this sort, the maximum value occurs $n-1$ times and we have reached the maximum.


If you want an explicit formula for the maximal value of the geometric mean, then you need to actually solve the equations $$\begin{matrix} y+(n-1) z &=& P \\ y^{-1} + (n-1) ^{-1} &=& Q \end{matrix}$$ This isn't too bad: Clearing denominators in the second equation and substituting in from the first gives $z+(n-1) (P - (n-1) z) = Q z(P-(n-1) z)$. This is a quadratic in $z$, so it is explicitly solvable, but I must say I don't find the solution illuminating. Anyway, take that solution and compute $y^{1/n} z^{(n-1)/n}$, and you will have computed the optimal value of the geometric mean when the arithmetic mean is $P/n$ and the harmonic is $n/Q$.

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I was flipping through old answers the other day, and I realized that we never resolved the original question. We showed that, given values for the AM and HM, the maximal value of the GM occurs when we take all but one of the $x_i$ to be equal, and the other to be less. In other words, we look at $x^{(n-1)/n} y^{1/n}$ where $x$ and $y$ are the roots of $$(1/n) x + (n-1)/n y = AM \quad 1/n x^{-1} + (n-1)/n y^{-1} = HM^{-1}$$ with $x>y$.

But we never worked out whether there was a nontrivial bound of the form $GM \leq \theta AM + (1-\theta) HM$.

The point of this answer is to point out that there is a nontrivial bound. But it's going to be pretty darn hard to find explicitly.

Following robjohn's excellent suggestion, take $\lambda=1/n$. More generally, we look at the equations $$AM= \lambda x + (1-\lambda) y \quad GM = x^{\lambda} y^{1-\lambda} \quad HM = (\lambda x^{-1} + (1-\lambda) y^{-1})^{-1}.$$ For any particular values of $(AM, GM, HM)$, the optimal value of $\theta$ is $\frac{GM-HM}{AM-HM}$. I found it most convenient to normalize $GM=1$, so $(x,y)$ are of the form $(e^{-(1-\lambda) t}, e^{\lambda t})$. What we want to show is that, for any $\lambda$, $$\sup_{t} \frac{GM-HM}{AM-HM} < 1.$$

Good news: The statement is true. Proof sketch: For $t \neq 0$, set $$f(\lambda, t) = \frac{GM-HM}{AM-HM} = \frac{1-(\lambda e^{(1-\lambda) t} + (1-\lambda) e^{- \lambda t})^{-1}}{(\lambda e^{-(1-\lambda) t} + (1-\lambda) e^{ \lambda t})- (\lambda e^{(1-\lambda) t} + (1-\lambda) e^{- \lambda t})^{-1}}.$$ A careful computation shows that $\lim_{t \to 0} f(\lambda, t) = 1/2$, so defining $f(\lambda, 0)=1/2$ gives a continuous function of $t$. Moreover, $f$ goes to $0$ as $t \to \pm \infty$. So the maximum of $f(\lambda, t)$ is obtained at some $t_0$. If $t_0$ were zero, then the max would be $1/2$. If $t_0$ were nonzero then $AM>GM$, so $\frac{GM-HM}{AM-HM} < 1$. Either way, the max is less than $1$. $\square$.

Bad news: Here is a plot of $f(1/3, t)$.

enter image description here

As you can see, the max does not occur at $t=0$. So we need to figure out where this max occurs in order to have a chance at it.

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