tofuu wrote: Real numbers $x_1, x_2, \dots, x_{2018}$ satisfy $x_i + x_j \geq (-1)^{i+j}$ for all $1 \leq i < j \leq 2018$. Find the minimum possible value of $\sum_{i=1}^{2018} ix_i$. $$ \sum \begin{Vmatrix} a & b\\c&d \end{Vmatrix}=a+b+c+d$$$$ \Rightarrow \sum \begin{Vmatrix} x_1 & x_2& x_3&...&x_{2018} \\ 0 & x_2 & x_3&...&x_{2018} \\ 0 & 0& x_3 &...&x_{2018}\\ ...& ...& ...& ...&....\\ 0& 0&0&...&x_{2018} \end{Vmatrix} = x_1 +2x_2+3x_3+...+2018x_{2018}=\sum_{i=1}^{2018} ix_i$$$$\sum_{i=1}^{2018} ix_i=\sum \left(\underbrace{\begin{Vmatrix} x_1 & x_2& x_3&...&x_{2018} \\ x_1 & x_2& x_3&...&x_{2018} \\x_1 & x_2& x_3&...&x_{2018} \\ ...& ...& ...& ...&....\\ x_1 & x_2& x_3&...&x_{2018} \end{Vmatrix}}_A -\underbrace{\begin{Vmatrix} 0 & 0& 0&...&0 \\ x_1& 0& 0&...&0\\x_1 & x_2& 0&...&0\\ ...& ...& ...& ...&....\\ x_1 & x_2& x_3&...&0 \end{Vmatrix}}_B \right)$$$$x_1+x_2+x_3+x_4+...+x_{2018}=\underbrace{x_1+x_2}+\underbrace{x_3+x_4}+...+\underbrace{x_{2017}+x_{2018}}\geq -1009 \Rightarrow \sum A\geq -1009\cdot 2018$$$$ \textbf{1)}x_1=x_1$$$$ \textbf{2)}x_1+x_2 \geq -1$$$$ \textbf{3)}x_1+x_2+x_3 \geq 1+x_2$$$$ \textbf{4)} x_1+x_2+x_3+x_4\geq 2$$$$\textbf{5)} x_1+x_2+x_3+x_4+x_5\geq 2+x_3$$$$.............................................................................$$$$ \textbf{2017)} x_{1}+x_{2}+...+x_{2017} \geq 1008+x_{1014}$$ $$\sum_{max} B \geq 2(2+3+...+1008)+\frac{2}{2}(x_1+x_2+x_3+...+x_{1009}) \geq 1010\cdot 1007+\frac{1}{2}\cdot1009$$$$ \sum_{i=1}^{2018} ix_i \geq \sum_{min} A-\sum_{max} B=-1009 \cdot 2018,5-1010\cdot 1007$$

My 200th post! Let $x_1=x_2=\cdots =x_{2018}=\frac {1}{2}$, then $\sum_{i=1}^{2018} ix_i=\frac {2037171}{2}$. We only need to prove $\sum_{i=1}^{2018} ix_i\geqslant \frac{2037171}2$. Lemma 1 For $n\geqslant 2$, $nx_n+(n+2)x_{n+2}+(n+4)x_{n+4}\geqslant\frac{3n+6}{2}=\frac{1}{2}[n+(n+1)+(n+2)]$. Lemma 1 proof $\begin{aligned}nx_n+(n+2)x_{n+2}+(n+4)x_{n+4}&=\left(\frac n2-1\right)(x_n+x_{n+2})+\left(\frac n2+3\right)(x_{n+2}+x_{n+4})+\left(\frac{n}{2}+1\right)(x_{n+4}+x_n)\\&\geqslant\left(\frac n2-1\right) +\left(\frac n2+3\right) +\left(\frac{n}{2}+1\right) =\frac{3n+6}{2}\end{aligned}$ Lemma 2 For $n\geqslant 1$, $nx_n+(n+2)x_{n+2}+(n+4)x_{n+4}+(n+6)x_{n+6}\geqslant 2n+6=\frac 12[n+(n+1)+(n+2)+(n+3)]$. Lemma 2 proof $\begin{aligned}nx_n+(n+2)x_{n+2}+(n+4)x_{n+4}+(n+6)x_{n+6}&=n(x_n+x_{n+2})+(n+4)(x_{n+4}+x_{n+6})+2(x_{n+2}+x_{n+6})\\&\geqslant n+n+4+2=2n+6\end{aligned}$ Proof As $1009=3\times 3+4\times 250$, using lemma 1 and lemma 2, we have $x_1+3x_3+\cdots +2017x_{2017}\geqslant \frac 12(1+3+\cdots +2017)$ $\cdots (*)$ $2x_2+4x_4+\cdots +2018x_{2018}\geqslant\frac 12(2+4+\cdots +2018)$ $\cdots (**)$ Add up $(*)$ and $(**)$ and we are done.