it's a problem from mine :
Let $x_i$ be $n$ positive real numbers such that $x_1\leq x_2\leq x_3\leq \cdots\leq x_{n-1}\leq x_{n}$ and $x_i-x_{i-1}=\alpha=constant$ where $2\leq i\leq n$ and finally $n\geq3$ then we have : $$\frac{n}{2}\Big(\Big(\prod_{i=1}^{n}x_i^{x_i}\Big)^{\frac{1}{\sum_{i=1}^{n}x_i}}+\Big(\prod_{i=1}^{n}x_i\Big)^{\frac{1}{n}}\Big)\leq \sum_{i=1}^{n}x_i$$
In fact it's a generalization of my last post Nice Inequality $2^23^34^45^5\cdots n^n \leq \Big(n+1-(n!)^{\frac{1}{n}}\Big)^{(n(n+1))0.5}$
The case $n=3$
$$\frac{3}{2}\Big(\Big(x_1^{x_1}x_2^{x_2}x_3^{x_3}\Big)^{\frac{1}{x_1+x_2+x_3}}+\Big(x_1x_2x_3\Big)^{\frac{1}{3}}\Big)\leq x_1+x_2+x_3$$
Or ($\frac{x_1+x_3}{2}=x_2$)
$$\frac{3}{2}\Big(\Big(x_1^{x_1}\frac{x_1+x_3}{2}^{\frac{x_1+x_3}{2}}x_3^{x_3}\Big)^{\frac{1}{x_1+\frac{x_1+x_3}{2}+x_3}}+\Big(x_1(\frac{x_1+x_3}{2})x_3\Big)^{\frac{1}{3}}\Big)\leq \frac{3}{2}(x_1+x_3)$$
Since it's homogeneous we can put $\frac{x_3}{x_1}=y\geq 1$ and get :
$$\frac{3}{2}\Big(\Big(\frac{y+1}{2}^{\frac{y+1}{2}}y^{y}\Big)^{\frac{1}{1+\frac{1+y}{2}+y}}+\Big((\frac{1+y}{2})y\Big)^{\frac{1}{3}}\Big)\leq \frac{3}{2}(1+y)$$
Wich is true graphically speaking .
My question :
Have you a method for the general case ?
Thank a lots for sharing your time and knowledge .