Following some ways on convexity I find this refinement of Am-Gm
Let $x_i>0$ be $n$ positives reals numbers then we have : $$\frac{\sum_{i=1}^{n}x_i}{n}\geq \Big(\Big(\prod_{i=1}^{n}x_i^{x_i}\Big)^{\frac{1}{\sum_{i=1}^{n}x_i}}\Big(\prod_{i=1}^{n}x_i\Big)\Big)^{\frac{1}{n+1}}\geq \Big(\prod_{i=1}^{n}x_i\Big)^{\frac{1}{n}}$$
I can prove the case $n=2$ :
Since the inequality is homogeneous (we can put $y=1$) we have to prove : $$x+1\geq 2\Big(\Big(x^x\Big)^{\frac{1}{x+1}}x\Big)^{\frac{1}{3}}$$
Wich can kills with power series .
For the general case I remember the proof of G.Pólya for classical Am-Gm maybe if we combine this wonderfull proof with power series the result comes .
If you have hints it would be nice .
Thanks.
For the left inequality.
Suppose $f:(0,\infty)\to \Bbb R$ with $f''(x)<0$ for all $x>0.$ If $x_1,...x_n$ are positive and if $W_1,...., W_n$ are non-negative with $\sum_{j=1}^nW_j=1$ then $$f\left(\sum_{j=1}^nW_jx_j\right)\ge \sum_{j=1}^nW_jf(x_j).$$
Geometrically this means that in the Cartesian plane, any point in the convex hull of $\{(x_1,f(x_1)),...,(x_n,f(x_n))\}$ lies "below" or on the graph of $f.$
Take the logarithms of the far-left and the middle terms in the Q and apply this with $f(x)=\ln x.$