It is not difficult to prove that if $x,y\in\mathbb{R}^+$ the inequality $$ \frac{x+y}{2}+\frac{2}{\frac{1}{x}+\frac{1}{y}}\geq \color{purple}{2}\cdot\sqrt{xy} $$ holds, and the constant $\color{purple}{2}$ is optimal.

In a recent question I proved, with a quite involved technique, that if $x,y,z\in\mathbb{R}^+$ then $$ \frac{x+y+z}{3}+\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\geq \color{purple}{\frac{5}{2\sqrt[3]{2}}}\cdot\sqrt[3]{xyz} $$ holds, and the constant $\color{purple}{\frac{5}{2\sqrt[3]{2}}}$ (that is a bit less than $2$) is optimal. Then I was wondering:

Given $x_1,x_2,\ldots,x_n\in\mathbb{R}^+$, what is the optimal constant $C_n$ such that: $$\text{AM}(x_1,\ldots,x_n)+\text{HM}(x_1,\ldots,x_n)\geq \color{purple}{C_n}\cdot \text{GM}(x_1,\ldots,x_n)$$

I do not think my approach with $3$ variables has a simple generalization (also because in $\mathbb{R}_+^3$ the stationary points are non-trivial), but maybe something is well-known about the improvements of the AM-GM inequality, or there is a cunning approach by some sort of induction on $n$.

Jack D'Aurizio
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    We can prove your general inequality by the (n-1) equal value method by Vasc. Link is here(http://math.stackexchange.com/questions/53853/combined-am-gm-qm-inequality) – HN_NH May 31 '16 at 23:46
  • @HN_NH: very nice, so this paper (http://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf) plus Henry's argument are enough to find $C_n$. Thank you guys! – Jack D'Aurizio May 31 '16 at 23:49

1 Answers1


Empirically it seems that the $n$ terms $\left(1,1,\ldots,1,1,\dfrac{1}{(n-1)^2}\right)$ suggest a low value for $C_n$ of $(n-1)^{2/n}\left(1+(n-1)^{-2}\right)$.

If $n-1$ of the terms are equal (and without loss of generality equal to $1$) and the other term is $x$ then $\frac{AM+HM}{GM}=\frac{\frac{x+n-1}{n}+\frac{n}{\frac{1}{x}+n-1}}{{{x}^{\frac{1}{n}}}}$ and its derivative is $$\frac{\left( n-1\right) \cdot {{\left( x-1\right) }^{2}}\cdot {{x}^{-\frac{1}{n}-1}}\cdot \left( {{n}^{2}}\cdot x-2\cdot n\cdot x+x-1\right) }{{{n}^{2}}\cdot {{\left( n\cdot x-x+1\right) }^{2}}}$$ which has zeros at $x=0,1,\frac{1}{(n-1)^2}$. Of these $x=0$ causes problems for the $GM$ and $HM$, while $x=1$ gives a value of $\frac{AM+HM}{GM}=2$ and a second derivative of $0$, and the interesting $x=\frac{1}{(n-1)^2}$ gives the value stated above which is less than $2$ and has a positive second derivative for $n\gt 2$

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  • Are you able to prove the given one is a stationary point? That would be quite a substantial step towards the solution. – Jack D'Aurizio May 31 '16 at 19:18
  • No - this is an empirical result - but it should not be difficult to find the optimal final term if all the others are equal – Henry May 31 '16 at 19:24
  • And indeed it is demonstrably correct when $n-1$ terms are equal – Henry May 31 '16 at 20:50