IMO 2019 Shortlisted Problems

Let $P=\left\{i:u_i>0\right\}$ and $N=\left\{i:u_i⩽0\right\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition ${\sum }_{i=1}^{2019}{u}_i=0$ we have $0={\sum }_{i=1}^{2019}{u}_i={\sum }_{i\in P}{u}_i-{\sum }_{i\in N}|u_i|$, so $$\begin{array}{c}\sum _{i\in P}{u}_i=\sum _{i\in N}|u_i|.\end{array}\text{\hspace{1em}(1)}$$ After this preparation, estimate the sum of squares of the positive and nonpositive elements as follows: $$\begin{array}{rl}& \sum _{i\in P}{u}_i^2⩽\sum _{i\in P}b{u}_i=b\sum _{i\in P}{u}_i=b\sum _{i\in N}|u_i|⩽b\sum _{i\in N}|a|=-nab;\\ & \sum _{i\in N}{u}_i^2⩽\sum _{i\in N}|a|\cdot |u_i|=|a|\sum _{i\in N}|u_i|=|a|\sum _{i\in P}{u}_i⩽|a|\sum _{i\in P}b=-pab.\end{array}\text{\hspace{1em}(2)(3)}$$ The sum of these estimates is $$1=\sum _{i=1}^{2019}{u}_i^2=\sum _{i\in P}{u}_i^2+\sum _{i\in N}{u}_i^2⩽-(p+n)ab=-2019ab;$$ that proves $ab⩽\frac{-1}{2019}$.

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Alternative solution.

As in the previous solution we conclude that $a<0$ and $b>0$. For every index $i$, the number $u_i$ is a convex combination of $a$ and $b$, so $$u_i=x_{i}a+y_{i}b\text{with some weights}0⩽x_i,y_i⩽1,\text{with}{x}_i+y_i=1.$$ Let $X={\sum }_{i=1}^{2019}{x}_i$ and $Y={\sum }_{i=1}^{2019}{y}_i$. From $0={\sum }_{i=1}^{2019}{u}_i={\sum }_{i=1}^{2019}\left(x_{i}a+y_{i}b\right)=-|a|X+bY$, we get $$\begin{array}{c}|a|X=bY.\end{array}\text{\hspace{1em}(4)}$$ From ${\sum }_{i=1}^{2019}\left(x_i+y_i\right)=2019$ we have $$\begin{array}{c}X+Y=2019\end{array}\text{\hspace{1em}(5)}$$ The system of linear equations $(4,5)$ has a unique solution: $$X=\frac{2019b}{|a|+b},Y=\frac{2019|a|}{|a|+b}$$ Now apply the following estimate to every $u_i^2$ in their sum: $$u_i^2=x_i^2{a}^2+2x_i{y}_{i}ab+y_i^2{b}^2⩽x_i{a}^2+y_i{b}^2$$ we obtain that $$1=\sum _{i=1}^{2019}{u}_i^2⩽\sum _{i=1}^{2019}\left(x_i{a}^2+y_i{b}^2\right)=X{a}^2+Y{b}^2=\frac{2019b}{|a|+b}|a{|}^2+\frac{2019|a|}{|a|+b}{b}^2=2019|a|b=-2019ab$$ Hence, $ab⩽\frac{-1}{2019}$.