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 $$S_h=\frac{(h-1)h^{\frac{1}{h-1}}}{e\log h}$$


1 Answer 1


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.

  • $\begingroup$ +1, but do you have a more elegant proof? $\endgroup$
    – Fischer
    Jun 1, 2013 at 0:02

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