TITLE: How to show $\frac{a_1+\cdots+a_n}{n}\le S_h\sqrt[n]{a_1\cdots a_n}$
QUESTION [5 upvotes]: For positive numbers $a_i$ with $0<m\le a_i\le M$, how to show 
$$\frac{a_1+\cdots+a_n}{n}\le S_h\sqrt[n]{a_1\cdots a_n},$$ where $h=M/m$ and $$S_h=\frac{(h-1)h^{\frac{1}{h-1}}}{e\log h}$$

REPLY [4 votes]: Let 
$$
f(b_1,\dots,b_n)=\frac{b_1^n+\dots+b_n^n}{nb_1\dots b_n},\quad \text{for $b_i\in [\sqrt[n]{m}, \sqrt[n]{M}]$}
$$
$f$ is a convex function for each variable, therefore the maximal is attained on the boundary
$$
\max f=\max_{s+t=n} \frac{sm+tM}{n \sqrt[n]{m^s}\sqrt[n]{M^t}}\le \max_{r\in [0, 1]} rh^{r-1}+(1-r)h^r.
$$
It remains to consider the function $g(r)=rh^{r-1}+(1-r)h^r,\, r\in [0, 1]$. 
The following is done with the help of http://www.wolframalpha.com/ (h=3, h=4, h=5, h=6). Since 
$$g'(r)=h^{r-1}\big[(-hr+h+r) \log(h)-h+1 \big],$$ 
put $g'(r)=0$, we obtain 
$$
\max g=S_h,
$$
completing the proof.