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?
 A: Looks like this is false for $n=3$.
$a_1,a_2,a_3=0.411022063900500, 0.438000608972404, 0.0731493447058247$.
A: 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. 

A: 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:



The following  link gives the picture using W/A : link

[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]

