AM-GM-HM Triplets

Question

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 $n=3$.

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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 $H = G^2$.

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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):

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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.*

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†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.]

Answer 1

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 $\mathbf{=1}$:

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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.

Answer 2

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.

Answer 3

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.

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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 $n/Q$.

Answer 4

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 $f(1/3, t)$.

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.

Answer 5

$\newcommand{\tr}{\operatorname{Tr}}$

Just came across this fun looking problem! Hopefully it's not too late to contribute something. Thought I'd add a few observations that don't directly resolve the original conjecture, but nonetheless explicitly upper bound the geometric mean by a (nontrivial) function of the arithmetic and harmonic means.

My argument is an information-theoretic one; it invokes some deep results, but a by-product is that the issue of relating the AM-GM and GM-HM inequalities can be viewed in a much more general context. To see what I mean by this, let $\gamma_s$ denote the centered Gaussian probability measure on $\mathbb{R}^n$ with covariance matrix $sI$, and let $\mu$ denote any other centered probability measure. Gross' logarithmic Sobolev (LSI) and Talagrand's inequalities are, respectively \begin{align} s I(\mu | \gamma_s) \geq 2 H(\mu | \gamma_s) \geq \frac{1}{s}W(\mu,\gamma_s)^2, ~~~~~(1) \end{align} where $I$ denotes relative Fisher information, $H$ denotes relative entropy, and $W$ is the quadratic Wasserstein distance. If we consider $\mu$ to be the centered Gaussian probability measure with covariance matrix $\Sigma$, then the LSI reduces (after algebra) to: \begin{align} s \frac{\tr(\Sigma^{-1})}{n} - \log s -1 \geq \frac{1}{n}\log \det(\Sigma^{-1}) ~~~\forall s>0.\notag \end{align} Minimizing the LHS over $s>0$ gives the inequality $\frac{1}{n}\tr(\Sigma^{-1}) \geq (\det (\Sigma^{-1}))^{1/n}$, which is precisely the AM-GM inequality applied to the eigenvalues of $\Sigma^{-1}$, which we'll denote henceforth by $\lambda_1, \lambda_2, \dots, \lambda_n$.

Next, Talagrand's inequality (after algebra) reduces to \begin{align} \frac{1}{n}\log \det(\Sigma^{-1}) \geq - 2\left(\frac{1}{\sqrt{s}} \frac{\tr(\Sigma^{1/2})}{n} - \log\frac{1}{\sqrt{s}} - 1 \right)~~~\forall s>0.\notag \end{align} Maximizing the RHS over $s>0$ gives the inequality $(\det (\Sigma^{-1}))^{1/n}\geq \left( \frac{n}{\tr(\Sigma^{1/2})}\right)^2$. The latter quantity is the squared Harmonic mean of $\lambda^{1/2}_1, \lambda^{1/2}_2, \dots, \lambda^{1/2}_n$, which is greater than the Harmonic mean of $\lambda_1, \lambda_2, \dots, \lambda_n$ by Jensen's inequality.

Therefore, when applied to a Gaussian measure $\mu$, the inequalities (1) capture the content of the AM-GM and GM-HM inequalities, respectively.

Now, it is known that the LSI and Talagrand inequalities do not exist in isolation. They are famously related by the so-called HWI inequality, which in the present setting is as follows: \begin{align} H(\mu | \gamma_s) \leq W(\mu, \gamma_s)\sqrt{I(\mu | \gamma_s)} - \frac{1}{2s}W(\mu,\gamma_s)^2.\notag \end{align} By the quadratic formula, we must have \begin{align} \sqrt{s I(\mu | \gamma_s)-2 H(\mu | \gamma_s)} \geq \sqrt{sI(\mu | \gamma_s) } - \frac{1}{\sqrt{s}}W(\mu,\gamma_s).\notag \end{align} To simplify notation, let me write $A$, $G$ and $H$ to denote the AM, GM and HM of $\lambda_1, \lambda_2, \dots, \lambda_n$. Further, let $H_{1/2}$ denote the HM of $\sqrt{\lambda_1}, \sqrt{\lambda_2}, \dots, \sqrt{\lambda_n}$. Then, particularizing the above general inequality to our setting gives

\begin{align} \left( s A - \log s -1 - \log G \right)^{1/2} \geq \left(\frac{1}{sH} -2 + s A \right)^{1/2} - \left( \frac{1}{sH} +1-\frac{2}{\sqrt{s H^2_{1/2}}} \right)^{1/2} ~~~\forall s>0. \label{HGA} \end{align}
Note that, for every $s>0$, this gives an interesting (dimension-free) upper bound on $G$ in terms of $A, H$ and $H_{1/2}$. This seems to provide an answer that speaks to the basic spirit of the conjecture.

Disclaimer: I would guess that the final inequality could be proved directly, now that the form is known. However, I have not tried to do so.