# Inequality of arithmetic, geometric and harmonic means

Let $a_1,\dots,a_n$ be positive numbers, does the following inequality holds? $$\frac{a_1+a_2+\cdots+a_n}{n}-\sqrt[n]{a_1a_2\cdots a_n}\geq\sqrt[n]{a_1a_2\cdots a_n}-\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$$ For $n=2$, it is trivial. For $n\geq3$, some numerical computation suggests that the inequality also holds. Could anyone prove it or disprove it?

• Over at math.SE, we have had two questions about for which $\theta$ (as a function of $n$), we have $GM \leq (1-\theta) AM + \theta HM$. (Your question asks whether $\theta=1/2$ works.) We know that the optimal $\theta$ goes to $0$ as $n \to \infty$, and $\theta=1/n$ works, but there is still some room for improvement. math.SE links: math.stackexchange.com/questions/92935 math.stackexchange.com/questions/803960 Sep 10, 2014 at 14:36
• Thank you very much for your answer. It's my first time to see this. Thank you again!
– HGF
Sep 10, 2014 at 14:41

Supplementing the picture of Aaron I've made a similar one, only, that I take into account, that the values $a,b,c$ are by definition greater than zero. Here I replaced them by exponentials of real values of the full real line; so if Aaaron's variables are finally $a_1= \frac ac, b_1= \frac bc, c_1=1$ I use $a_1=\exp(x)$ and $b_1=\exp(y)$ and $c_1=\exp(0)$.
For the plot I ask then for the sign $$f(x,y)= \operatorname{sign} \left( {1+e^x+e^y\over3}+{3 \over 1+e^{-x}+e^{-y}} - 2 \cdot \exp\left({x+y \over 3}\right)\right)$$. Heuristically I needed only the range $-2 \cdots 2$ for the x and y-parameters. The color is green where $f(x,y)=1$ (thus the original inequality holds, and the color is white , where $f(x,y)=-1$. I don't know, whether $f(x,y)=0$ really exists. Here is the image:

[update] A short step into the generalization to n-tuples.
Once we have the replacement of the definition of the tuples $(a_1,a_2,a_3,....a_n)$ by that of the tuples $(e^{b_1},e^{b_2},e^{b_3},...,e^{b_n})$ we have $${e^{b_1}+e^{b_2}+e^{b_3}+...+e^{b_n} \over n} + {n \over e^{-b_1}+e^{-b_2}+e^{-b_3}+...+e^{-b_n} } \ge 2\cdot e^{{b_1+b_2+b_3+...+b_n \over n}} \tag 1$$ If we now define $M={\sum_{k=1}^n b_k \over n}$ as mean of the coefficients $b_k$ and then the coefficients $c_k=b_k-M$ then we can simplify: $$e^M \cdot {e^{c_1}+e^{c_2}+e^{c_3}+...+e^{c_n} \over n} + e^M \cdot {n \over e^{-c_1}+e^{-c_2}+e^{-c_3}+...+e^{-c_n} } \ge 2\cdot e^M \tag 2$$ and finally $${e^{c_1}+e^{c_2}+e^{c_3}+...+e^{c_n} \over n} + {n \over e^{-c_1}+e^{-c_2}+e^{-c_3}+...+e^{-c_n} } \ge 2 \tag 3$$ We expand now the exponential-expressions into their power series. Then the constant and linear terms vanish and we get an inequality of series which begin with the quadratic terms.
We define for notational convenience: $$t_k = {c_1^k+c_2^k+c_3^k+...+c_n^k \over k!}$$ and $$w_k^+ = t_k + t_{k+1} + t_{k+2} + t_{k+3} + t_{k+4} + ... \\ w_k^- = t_k - t_{k+1} + t_{k+2} - t_{k+3} + t_{k+4} - ... \\$$ then the first numerator in (3) is $$e^{c_1}+e^{c_2}+e^{c_3}+...+e^{c_n} = n + t_1 + w_2^+ \tag 4$$ and the second denominator in (3) is $$e^{-c_1}+e^{-c_2}+e^{-c_3}+...+e^{-c_n} = n - t_1 + w_2^- \tag 5$$ By definition of the coefficients $c_k$ we have moreover $t_1 = 0$ and inequality (3) becomes $${n + w_2^+ \over n} + {n \over n + w_2^- } \ge 2 \tag 6$$ then $${ w_2^+ \over n} -{ w_2^- \over n} \cdot {1 \over {1+ w_2^- \over n}} \ge 0 \tag 7$$ and finally $$w_2^+ \ge {n\cdot w_2^- \over 1+ w_2^- } \tag 8$$

The multidimensional generalization of the plot/the figure above would now be an interesting exercise, but unfortunately I cannot at the moment analyze this furtherly. [/update]

• $f(x,y)=0$ exists. Fix $y$ and plot the function without sign of $x$, there are sign changes in the plot.
– joro
Sep 11, 2014 at 14:34
• @joro: true. Everything is continuous and a sign-change occurs, so a zero must occur. (I just was in a hurry and didn't want to write something wrong...) Sep 11, 2014 at 20:58
• In my graph, if extended to the whole (strictly) positive quadrant, one would have the symmetries $(x,y) \mapsto (y,x)$ and $(x,y) \mapsto (1/x,y/x).$ Together these give six symmetries. With your parameterization the first is the same but the second becomes $(x,y) \mapsto (-x,y-x).$ Similar symmetries make me suspect that, for $d+1 \gt 2$ variables, your graph will have $d+1$ finite lobes in $\mathbb{R}^d$ going from the origin. One centered on the line $x_1=x_2=\cdots=x_{d-1} \gt 0$ and the rest on the $d-1$ negatives axes. Sep 12, 2014 at 6:30
• @Aaron: yes, I think that in the d-dimensional case it is just like this. I've tried to confirm the intuition for the 3-d case, drawing the pictures with an additional variable $z$ with layers defined by constant values for $z$. It comes out to be as expected, only the shape of the "leaves" changes: they become longer and more slim. I'd like to find an estimate for the extension on the negative axes. Sep 12, 2014 at 8:14
• I'm not sure that the exponentials help. Set all the valuese but one to $1=e^0$ and leave the last a variable. A nice trick is to set it to $x^d$ (or $e^{dz}$ if you insist). Then $A+H-2G$ is a rational function of $x$. The numerator has a factor of $(x-1)^d$ and you can look at the roots of the other factor. Remember to raise to the $d$th power. That is actually not the local minimum although it is the fixed point of a symmetry. Sep 12, 2014 at 8:37

Looks like this is false for $n=3$.

$a_1,a_2,a_3=0.411022063900500, 0.438000608972404, 0.0731493447058247$.

• Oh, God. Thank you very much for your counter example
– HGF
Sep 10, 2014 at 14:29
• @user48985 it fails for $n=1$ too.
– joro
Sep 10, 2014 at 14:54
• Oh! I think it is right for n=1, it became $0\geq 0$.
– HGF
Sep 10, 2014 at 15:30
• @user48985 You appear right, I didn't verify correctly $n=1$.
– joro
Sep 10, 2014 at 15:39

Since it is homogeneous you can assume that $c=1.$ The graph below on the right shows the region where the inequality fails. The graph on the left is just the restriction to $a,b \le 2$ (for $c=1$).

The map $(a,b) \mapsto (\frac{1}{a},\frac{b}{a})$ rotates the three lobes clockwise. Along with reflection in the line $a=b$ this gives six symmetries. The big lobe goes out to a point $P=(m,m)$ where $m\approx 6.638.$ Some calculation shows that the exact value is the largest root of $t^3-3t^2-24t-1$. That point is mapped by $(a,b) \mapsto (\frac{b}{a},\frac{1}{a})$ to the point $Q=(1,\frac{1}{m}).$ It is almost, but not quite, the lowest point of the region. The tangent line $a+b=2m$ at $P$ maps to a line of slope about $0.075$ which is the tangent at $Q$. Here is a plot of that region.

• You mean the dark part of this region?
– HGF
Sep 11, 2014 at 3:51
• Correct. So in ratio roughly 1/4:1:1 we might take $a=125=5^3$ and $b=c=512=8^3.$ Then the arithmetic,geometric, and harmonic means are $A=383$,$G=320$ and $H=\frac{125\ 256}{127} \approx 251.9685 \lt 252.$ so $A-G=63 \lt 68 \lt G-H.$ Sep 12, 2014 at 4:11
• I've found another representation for the value of around $m \approx 6.64$: let $w$ be the cubic root of complex unity $w=1/2 - 0.866... î$ then $m = (w^{1/3}+w^{-1/3})^3$ (in my answer I had just $x=\log(m)$ ) Sep 12, 2014 at 10:00
• An even more general solution for the value m for the tuple of $d-1$ is: let $\rho_d$ be the root of $g_d(x)= x \cdot ({x^{-d}+d-1 \over d}+{d \over x^d+d-1})-2$ then simply $m_d=\rho_d^d$. It gives for $m_3 \approx 6.638$ and $m_4 \approx 22.59$ and so on. Sep 12, 2014 at 10:40