$(x_1+x_2+\cdots+x_{2023})^2-(x_1^2+3x_2^2+5x_3^2+\cdots+(2\cdot2023-1)\cdot x^2_{2023})=2 \sum_{i=2}^{2023} x_i( \sum_{j=1}^{i-1} x_j-(i-1)x_i) \geq 0$ because $ \sum_{j=1}^{i-1} x_j-(i-1)x_i \geq 0$ And equality if all $x_i$ are equal. Or we can prove it by induction using $(x_1+x_2+...+x_n)^2-(x_1^2+3x_2^2+5x_3^2+\cdots+(2\cdot n-1)\cdot x^2_{n})=(x_1+x_2+...+x_{n-1})^2-(x_1^2+3x_2^2+5x_3^2+\cdots+(2\cdot (n-1)-1)\cdot x^2_{n-1})+x_n^2+2x_n(x_1+\cdots +x_{n-1})-(2\cdot n-1)\cdot x^2_{n}=(x_1+x_2+...+x_{n-1})^2-(x_1^2+3x_2^2+5x_3^2+\cdots+(2\cdot (n-1)-1)\cdot x^2_{n-1})+2x_n(x_1+\cdots +x_{n-1}-( n-1)\cdot x_{n}) \geq 0$

Actually Equality can be when some are 0, and the others are equal.