About an inequality including arithmetic mean, geometric mean and harmonic mean 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?
 A: By Maclaurin's Inequality, $S_1 \ge \sqrt[n-1]{S_{n-1}} \implies S_1^{n-1} \ge S_{n-1}  \implies A^{n-1}H \ge G^n$.  
So if we normalise with $A=1$, we know $0 < H \le G \le A = 1$ and we have $H \ge G^n$ with equality when $H=G=A=1$.
The normalised inequality is $p+(1-p)H \ge G$, and with the above bound, it is enough to have $p+(1-p)G^n \ge G$.  As this must hold $\forall G \in (0, 1]$, we must have:
$$p \ge \frac{G-G^n}{1-G^n}$$
As the RHS achieves a maximum of $1-\frac1n$ when $G \to 1$, we need $p_n \ge 1-\frac1n$.  Thus we have:$$\left(1-\tfrac1n \right)A+\tfrac1nH\ge G$$

Updated based on David Speyer's comment and post, as an upper bound rather than the optimal $p_n$ ...
For $n=3, 4, 5, ...10$, the minimal $p_n$ seem to be $\approx$
\begin{array}{ l | c }
\hline
  3 & 0.52605 \\ \hline
  4 & 0.56301 \\ \hline
  5 & 0.59660 \\ \hline
  6 & 0.62560 \\ \hline
  7 & 0.65055 \\ \hline
  8 & 0.67213 \\ \hline
  9 & 0.69095 \\ \hline
  10 & 0.70752 \\ \hline
  \hline
\end{array}
P.S.  The table above was calculated by numerically finding the maximum of 
$$p_n = \max_{x>1} \frac{G_n-H_n}{1-H_n} = \max_{x>1} \frac{\sqrt[n]{x^{n-1}(n-(n-1)x)}-\frac{n}{(n-1)/x+1/(n-(n-1)x)}}{1-\frac{n}{(n-1)/x+1/(n-(n-1)x)}}$$
A: 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.
