TITLE: About an inequality including arithmetic mean, geometric mean and harmonic mean
QUESTION [17 upvotes]: For any $n$ positive real numbers $a_i\ (i=1,2,\cdots,n)$, let us define $A,G,H$ as
$$A=\frac{\sum_{i=1}^{n}a_i}{n},\ G=\sqrt[n]{\prod_{i=1}^{n}a_i},\ H=\frac{n}{\sum_{i=1}^{n}\frac{1}{a_i}}.$$ 
Then, here is my question.

Question : How can we find the minimum value of a real number $p$ which satisfies the following inequality for any $a_i(i=1,2,\cdots,n)$?
  $$pA+(1-p)H\ge G$$

I've already found that the answer for $n=2$ is $p=1/2$. However, I haven't had any good idea for $n\ge 3$ in general. Can anyone help?

REPLY [5 votes]: In this answer, it is shown that for $n$ items with an arithmetic mean of $1$ and a harmonic mean of $h$, the geometric mean is between $g(h,\frac1n)$ and $g(h,\frac{n-1}n)$ where
$$
g(h,\lambda)=\frac{\left(\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}+1\right)^\lambda\left(\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}-1\right)^{1-\lambda}}{\sqrt{1+4\frac{h}{1-h}\lambda(1-\lambda)}+2\lambda-1}\tag{1}
$$
We can compute the minimum $p$ that will work for a given $n$ and $h$ with
$$
p(n,h)=\frac{g(h,\frac{n-1}{n})-h}{1-h}\tag{2}
$$
Maximizing $p(n,h)$ for $h\in[0,1]$, we get
$$
\begin{array}{r|c|c}
n&p&h\\\hline
2&0.50000000&1.00000000\\\hline
3&0.52604991&0.83027796\\\hline
4&0.56301305&0.67404567\\\hline
5&0.59659653&0.57799531\\\hline
6&0.62560358&0.51531972\\\hline
7&0.65054843&0.47136304\\\hline
8&0.67212783&0.43872876\\\hline
9&0.69095251&0.41342714\\\hline
10&0.70751470&0.39314442\\\hline
\end{array}\tag{3}
$$
which agrees with Macavity's answer.