# How find this maximum $\sum_{i=1}^{2000}\left(\frac{x^{2000}_{i}}{\sum_{j=1}^{2000}x^{3999}_{j}- i\cdot x^{3999}_{i}+2000}\right)$

Question:

let $x_{1},x_{2},\cdots,x_{2000}$ be real numbers,and such $x_{i}\in [0,1],i=1,2,\cdots,2000$.and define $$F_{i}=\dfrac{x^{2000}_{i}}{\displaystyle\sum_{j=1}^{2000}x^{3999}_{j}- \textbf{i }\cdot x^{3999}_{i}+2000}$$

Find the maximum of the $$\sum_{i=1}^{2000}F_{i}$$

before when I solve this problem,I have solve it follow problem:

let $x_{1},x_{2},\cdots,x_{2000}$ be real numbers,and such $x_{i}\in [0,1],i=1,2,\cdots,2000$.and define $$F_{i}=\dfrac{x^{2000}_{i}}{\sum_{j=1}^{2000}x^{3999}_{j}-x^{3999}_{i}+2000}$$

Find the maximum of the $$\sum_{i=1}^{2000}F_{i}$$

Because $x_{i}\in[0,1]$, so It is clear $$\sum_{i=1}^{2000}F_{i}\le\dfrac{\sum_{i=1}^{2000}x^{2000}_{i}}{\displaystyle\sum_{i=1}^{2000}x^{3999}_{i}+1999}=S$$ Indeed,we can prove $$S\le\dfrac{2000}{3999}$$ or $$\sum_{i=1}^{2000}x^{2000}_{i}\le\dfrac{2000}{3999}(\sum_{i=1}^{2000}x^{3999}_{i}+1999)$$ because we can easy to prove $$\dfrac{2000}{3999}x^{3999}_{i}-x^{2000}_{i}+\dfrac{1999}{3999}\ge 0,x_{i}\in[0,1]$$

when $x_{1}=x_{2}=\cdots=x_{2000}=1$ But my problem is not this.

so I can't prove it. Thank you