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?

8$\begingroup$ 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 $\endgroup$– David E SpeyerSep 10, 2014 at 14:36

$\begingroup$ Thank you very much for your answer. It's my first time to see this. Thank you again! $\endgroup$– HGFSep 10, 2014 at 14:41
3 Answers
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 yparameters. 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:
The following link gives the picture using W/A : link
[update] A short step into the generalization to ntuples.
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 $c_k=b_kM$ 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 exponentialexpressions 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]

$\begingroup$ $f(x,y)=0$ exists. Fix $y$ and plot the function without sign of $x$, there are sign changes in the plot. $\endgroup$– joroSep 11, 2014 at 14:34

$\begingroup$ @joro: true. Everything is continuous and a signchange occurs, so a zero must occur. (I just was in a hurry and didn't want to write something wrong...) $\endgroup$ Sep 11, 2014 at 20:58

$\begingroup$ 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 $(x,y) \mapsto (x,yx).$ 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 $x_1=x_2=\cdots=x_{d1} \gt 0$ and the rest on the $d1$ negatives axes. $\endgroup$ Sep 12, 2014 at 6:30

$\begingroup$ @Aaron: yes, I think that in the ddimensional case it is just like this. I've tried to confirm the intuition for the 3d 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. $\endgroup$ Sep 12, 2014 at 8:14

$\begingroup$ 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 $A+H2G$ is a rational function of $x$. The numerator has a factor of $(x1)^d$ 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. $\endgroup$ Sep 12, 2014 at 8:37
Looks like this is false for $n=3$.
$a_1,a_2,a_3=0.411022063900500, 0.438000608972404, 0.0731493447058247$.

$\begingroup$ Oh, God. Thank you very much for your counter example $\endgroup$– HGFSep 10, 2014 at 14:29


$\begingroup$ Oh! I think it is right for n=1, it became $0\geq 0$. $\endgroup$– HGFSep 10, 2014 at 15:30

$\begingroup$ @user48985 You appear right, I didn't verify correctly $n=1$. $\endgroup$– joroSep 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 $t^33t^224t1$. 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.


$\begingroup$ 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 $AG=63 \lt 68 \lt GH.$ $\endgroup$ Sep 12, 2014 at 4:11

$\begingroup$ 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 $x=\log(m)$ ) $\endgroup$ Sep 12, 2014 at 10:00

$\begingroup$ An even more general solution for the value m for the tuple of $d1$ is: let $\rho_d$ be the root of $g_d(x)= x \cdot ({x^{d}+d1 \over d}+{d \over x^d+d1})2$ then simply $m_d=\rho_d^d$. It gives for $m_3 \approx 6.638$ and $m_4 \approx 22.59$ and so on. $\endgroup$ Sep 12, 2014 at 10:40