WLOG $u_1 \le \cdots \le u_{2019}$. Split these into the positives and negatives: let $u_1,\ldots,u_x$ be the negatives, and let $u_{x+1},\ldots,u_{x+y}$ be the nonnegatives, where $x+y=2019$. Let $a_1=-u_1,\ldots,a_x=-u_x$, and let $b_1=u_{x+1}, \ldots, b_y=u_{x+y}$. Rephrased problem wrote: Let $0\le a_1 \le \cdots \le a_x$ and $0\le b_1\le \cdots \le b_y$. We know $a_1+\cdots+a_x=b_1+\cdots+b_y$, and $a_1^2+\cdots+a_x^2+b_1^2+\cdots+b_y^2=1$. Prove $a_xb_y \ge \tfrac{1}{x+y}$. Let $S=a_1+\cdots+a_x=b_1+\cdots+b_y$. We know $S\le xa_x$ and $S\le yb_y$. So $b_yS \le xa_xb_y$ and $a_xS\le ya_xb_y$. Summing, $a_xS + b_yS \le (x+y)a_xb_y$. Now, we have \begin{align*} \frac{1}{x+y} &= \frac{a_1^2+\cdots+a_x^2+b_1^2+\cdots+b_y^2}{x+y} \\ &\le \frac{a_x(a_1+\cdots+a_x) + b_y(b_1+\cdots+b_y)}{x+y} \\ &= \frac{a_xS + b_yS}{x+y} \\ &\le \frac{(x+y)a_xb_y}{x+y} = a_xb_y. \end{align*}

https://artofproblemsolving.com/community/c6h222731p1236019 lol

y-is-the-best-_ wrote: Let $u_1, u_2, \dots, u_{2019}$ be real numbers satisfying \[u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.\]Let $a=\min \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$ and $b=\max \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$. Prove that \[ a b \leqslant-\frac{1}{2019}. \] We'll prove this statement for any constant $k$ rather than $2019$, furthermore we discard all the elements of $u_i$ which are zero. Now, we can divide $u_i$s into positive and negative: $a_1 \ge a_2 \ge \dots \ge a_m > 0 >a_{m + 1} \ge \dots \ge a_{m + n} $. Now, for all the negative variables, define $b_i = -a_{m + i}$. Let the set of $a_i$ has size $|\mathcal{P}|$ and the set of $b_i$ has size $|\mathcal{N}|$. We wanted to prove that \[ ab \ge \frac{1}{|\mathcal{P}| + |\mathcal{N}|} \]Notice that \[ 1 = \sum_{i = 1}^{|\mathcal{P}|} a_i^2 + \sum_{j = 1}^{|\mathcal{N}|} b_j^2 \le b \sum_{i = 1}^{|\mathcal{P}|} a_i + a \sum_{j = 1}^{|\mathcal{N}|} b_j = b \sum_{j = 1}^{|\mathcal{N}|} b_j + a \sum_{j = 1}^{|\mathcal{P}|} a_i \le ab |\mathcal{N}| + ab|\mathcal{P}| = ab ( |\mathcal{N}| + |\mathcal{P}|)\]We are done.

Since sum of the squares is $1$ not all $u_i$ are zero. Since the sum is $0$ some of these numbers must be positive and some negative. Let $-a_1 \leq -a_2 \leq \dots \leq -a_k < 0$ be all negative numbers among $u_i$ and let $ b_1 \geq b_2 \geq \dots \geq b_s \geq 0$ be all non-negative numbers among $u_i$. Then $a = - a_1$, $b = b_1$, $k+s = 2019$, $$a_1 + a_2 + \cdots + a_{k} = b_1 + b_2 + \cdots + b_s \quad\text{and}\quad a_1^2 + a_2^2 + \cdots + a_{k}^2 + b_1^2 + b_2^2 + \cdots + b_{s}^2 = 1$$Then $$a_1^2 + a_2^2 + \cdots + a_{k}^2 \leq a_1a_1 + a_1a_2 + \cdots + a_{1}a_k = a_1b_1 + a_1b_2 + \cdots + a_1b_s \leq sa_1b_1$$Similarly, $$b_1^2 + b_2^2 + \cdots + b_{s}^2 \leq b_1b_1 + b_1b_2 + \cdots + b_{1}b_s = b_1a_1 + b_1a_2 + \cdots + b_1a_k \leq ka_1b_1$$Adding these two inequalities side by side we get $$1 \leq (k+s)a_1b_1 = 2019a_1b_1 \Longrightarrow ab = -a_1b_1 \leq -\frac{1}{2019}$$

Comment: This problem requires at least one tricky algebra move. Hence, it's way too hard for A2. In my opinion this could be like 25-30M. Let $\{u_1,u_2,\hdots,u_N\} = \{a_1,a_2,\hdots,a_k,-b_1,-b_2,\hdots,-b_{\ell}\}$ where $a_1,\hdots,a_k,b_1,\hdots,b_{\ell}>0$. Assume for the contradiction that $ab>\tfrac{-1}{N}$, then $a_ib_j<\tfrac 1N$ for any $i,j$. Now we consider the sum $$T=\sum_{i=1}^k\sum_{j=1}^{\ell}a_ib_j(a_i+b_j)$$in two ways. First, we have \begin{align*} T &= (a_1+\hdots+a_k)(b_1^2+\hdots+b_{\ell}^2) + (b_1+\hdots+b_{\ell})(a_1^2+\hdots+a_k^2) \\ &= (a_1+\hdots+a_k)(a_1^2+\hdots+a_k^2+b_1^2+\hdots+b_{\ell}^2) \\ &= a_1+a_2+\hdots+a_k \end{align*}On the other hand, \begin{align*} T &< \frac 1N \sum_{i=1}^k\sum_{j=1}^{\ell}(a_i+b_j) \\ &= \frac{\ell(a_1+\hdots+a_k) + k(b_1+\hdots+b_{\ell})}{N} \\ &= a_1+a_2+\hdots+a_k \end{align*}which is a contradiction.

Due to $u_i \in [a, b]$ we have: \[1 + 2019ab = \sum_{i = 1}^{2019} u_i^2 -(a+b)\sum_{i = 1}^{2019} u_i + 2019ab = \sum_{i = 1}^{2019} (u_i - a)(u_i - b) \leqslant 0\]

Ainulindale wrote: https://artofproblemsolving.com/community/c6h222731p1236019 lol Fun fact: the source of this is... ISL 1972.

Cute problem. First note that not all the numbers are positive or all negative. Hence $a<0$ and $b>0$. Consider disjoint subsets $S$ and $T$ of $\{1,2,\dots ,2019\}$ such that $u_i>0$ for all $i \in S$ and $u_i<0$ for all $i \in T$ . Suppose $\vert S\vert= s$ and $\vert T \vert=t$. Note that : $$ \sum_{i=1}^{2019} u_i =0 \implies\sum_{i\in S}u_i = \sum_{j \in T} \vert u_j\vert$$ Next note that we have the estimates : $$\sum_{i\in S} u_i^2 \leq b\cdot \left (\sum_{i\in S} u_i \right) \leq b\cdot \left (\sum_{j\in T} \vert u_j \vert \right) \leq -abt$$$$\sum_{j\in T} u_j^2 \leq \vert a \vert \cdot \left (\sum_{j\in T} \vert u_i \vert \right) \leq b\cdot \left( \sum_{i\in S} u_i \right) \leq -abs$$ Summing up we have : $$ 1 = \sum_{i=1}^{2019} u_i^2 = \sum_{i\in S}u_i^2 + \sum_{j\in T} u_j^2 \leq -abt-abs = -2019ab$$ Hence we have $\boxed{ab \leq -\frac 1{2019}}$ as desired. $\blacksquare$

Everyone tends to overcomplicate this. Just use that $(u_i-a)(u_i-b) \le 0$ for $1\le i\le 2019$ and sum up all the indices.

a1267ab wrote: Fun fact: the source of this is... ISL 1972. Ahhh posted! Anyways will still post y-is-the-best-_ wrote: Let $u_1, u_2, \dots, u_{2019}$ be real numbers satisfying \[u_{1}+u_{2}+\cdots+u_{2019}=0 \quad \text { and } \quad u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1.\]Let $a=\min \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$ and $b=\max \left(u_{1}, u_{2}, \ldots, u_{2019}\right)$. Prove that \[ a b \leqslant-\frac{1}{2019}. \] Solution: (Lemma). Let $u_1 ,u_2 ,\ldots, u_n$ be real numbers satisfying $u_{1}+u_{2}+\cdots+u_{n}=0.$ If $a=\min \left(u_{1}, u_{2}, \cdots, u_{n}\right)$ and $b=\max \left(u_{1}, u_{2}, \cdots, u_{n}\right)$, then, $u_1^2 + u_2^2 +\cdots +u_n^2 \leqslant - nab.$ Proof: IMOSL 1972/3.$\square$ Here $n=2019$ and $u_{1}^{2}+u_{2}^{2}+\cdots+u_{2019}^{2}=1$$$\implies 1 \leq -2019ab \implies -\frac{1}{2019} \geqslant ab. \quad \blacksquare$$

Throw out all zero values of $u_i$, which is fine because there clearly exists nonzero $u_i$, which implies there exist both positive and negative $u_i$. Now, say there are $k$ positive $u_i$ and $j$ negative $u_i$. Let the $j$ negative $u_i$ be $a_1\ge a_2\ge\cdots\ge a_j<0$ and the $k$ positive $u_i$ be $-b_1\le -b_2\le\cdots\le -b_k>0$. Note that $\sum_i b_i = \sum_i a_i=S$, and \[\sum_i a_i^2+\sum_i b_i^2=1.\]We seek to show that $a_1b_1\ge \frac{\sum_i a_i^2+\sum_i b_i^2}{j+k}= \frac{1}{j+k}\ge \frac{1}{2019}$. Equivalently, we seek to show that \[(j+k)a_1b_1\ge \sum_i a_i^2+\sum_i b_i^2.\]Note that $\sum_i a_i=\sum_i b_i>0$. As the inequality is homogenous, we may assume $\sum_i a_i=\sum_i b_i=1$. Now, the following stupid approach works; note \[ja_1b_1 = \left(\underbrace{a_1+\cdots+a_1}{j}\right)b_1 \ge \left(\sum_i a_i\right)b_1 = b_1 = b_1\left(b_1+\cdots+b_n\right) = b_1^2+b_1b_2+\cdots+b_1b_n\ge \sum_i b_i^2,\]but we similarly have $ka_1b_1\ge \sum_i a_i^2$, hence we are done.

WLOG none of the terms are zero (because thats easy to deal with, even though i shouldnt be wlogging this lol). Let $a_1, \dots, a_k \in \{u_1, \dots, u_{2019}\}$ be all of the negative terms, and $\{b_1, \dots, b_n\} \in \{u_1, \dots, u_{2019}\}$ be all of the positive terms. Let $l$ and $m$ be the number of positive terms and negative terms respectively. We now want the minimal value of $\max(a_1^2, \dots, a_k^2)\max(b_1^2, \dots b_n^2).$ But note that $\frac{(\sum a_i^2 + \sum b_i^2)}{l+m} = \frac{1}{l+m} = \frac{1}{2019},$ so it is sufficient to prove $\sqrt{\max(a_1^2, \dots, a_k^2)\max(b_1^2, \dots b_n^2)} \ge \frac{(\sum a_i^2 + \sum b_i^2)}{l+m}.$ This is easy since $\sqrt{\max\{a_1^2, \dots, a_k^2\}\max\{b_1^2, \dots b_n^2\}} = \max\{|a_1|, \dots, |a_k|\}\max\{b_1, \dots b_n\} \ge \frac{(\sum a_i^2 + \sum b_i^2)}{l+m}$

Take $n=2019$ Wlog, assume that $u_1\le u_2\le u_3\dots\le u_n$.Let $X$ be a discrete random variable which takes value $u_1,u_2\dots u_n$ uniformly and randomly. At first observe that,\begin{align*} \mathbb E[(X-\frac{u_1+u_n}{2})^2]&=\frac{1}{n}\sum_i (u_i-\frac{u_1+u_n}{2})^2\\ &=\frac{1}{n}\sum_{i;u_i\le \frac{u_1+u_n}{2}}(u_i-\frac{u_1+u_n}{2})^2+\sum_{i;u_i>\frac{u_1+u_n}{2}}+(u_i-\frac{u_1+u_n}{2}))^2\\ &\le \frac{1}{n}\sum_{i;u_i\le \frac{u_1+u_n}{2}}(u_1-\frac{u_1+u_n}{2})^2+\sum_{i;u_i>\frac{u_1+u_n}{2}}((u_n-\frac{u_1+u_n}{2}))^2\\ &=\frac{1}{4} (u_n-u_1)^2 \end{align*}Next since $E(x^2)=\frac{1}{n}$ and $E(X)=0$,so, \begin{align*} \mathbb E[(X-\frac{u_1+u_n}{2})^2]= E[X^2]+(\frac{u_1+u_n}{2})^2=\frac{1}{n}+ (\frac{u_1+u_n}{2})^2 \end{align*}Combining these 2 we get , $$u_1u_n\le -\frac{1}{n}$$$\blacksquare$

By Karamata, we know that $1\leq ma^2+nb^2+c^2$ (The numbers of a is m and numbers of b is n)，such that $ma+nb+c=0,m+n=2018,a\leq c\leq b$ $1\leq ma^2+nb^2+(ma+nb)^2=(m^2+m)a^2+2mnab+(n^2+n)b^2$ $1+2019ab\leq \big(ma+(n+1)b\big)\big((m+1)a+nb\big)=(a-c)(b-c)$ $1+2018ab\leq -c(a+b-c)\leq -ab$ $ab\leq\frac{-1}{2019}$

For every $u_i$ : $(a-u_i)(b-u_i)\le 0$ thus $\sum_{i=1}^{n} (a-u_i)(b-u_i)=nab+1\le 0$ , simple as that . Also this is sharp for every $n$. Just set $u_1=u_2=...=u_{n-1}=\frac{1}{\sqrt{n(n-1)}}$ and $u_n=-(n-1)\frac{1}{\sqrt{n(n-1)}}$.

A different solution: Consider the quadratic function $$f(x)= \frac{1}{2019} \sum_{i=1}^{2019}(x-u_i)^2+(x-a)(x-b) = 2x^2-(a+b)x+\frac{1}{2019}+ab.$$We have to prove that $f(0) \leq 0$. But we know that every quadratic function is symmetric with respect to its extremum. Since $f'(x)=0$ is satisfied for $x=\frac{a+b}{4}$, we have in particular that $f(0)=f \left( \frac{a+b}{2} \right)$. Because $\left( \frac{a+b}{2} - u_i \right)^2 \leq \frac{(a-b)^2}{4}$ for each $i$, it follows that $$ f \left( \frac{a+b}{2} \right) = \frac{1}{2019} \sum_{i=1}^{2019}\left( \frac{a+b}{2} - u_i \right)^2- \frac{(a-b)^2}{4} \leq 0,$$as desired.

I used to think this is not hard but now I believe this is perfect for A2. Had a bit of trouble reconstruction the soution. We shall partition $u_1, u_2, \ldots , u_{2019}$ into two groups: $v_1 \to v_i$ and $w_1 \to w_{2019 - i}$, where the $v_i$'s are nonnegative and the $w_i$'s are negative. For each $j$ from $1 \to 2019 - i$ let $x_i = -w_i$ which is positive. We have\[v_1 + \ldots + v_{i} = x_1 + \ldots + x_{2019 - i} = K\]where $(v_1^2 + \ldots + v_{i}^2) + (x_1^2 + \ldots + x_{2019 - i}^2) = 1$ must hold. If we are to assume $v_i$ and $x_i$ are nonincreasing, which is allowed WLOG, then we want to show that $v_1x_1 \geq \tfrac{1}{2019}$. Note that by Cauchy-Schwarz,\[v_1^2 + v_2^2 + \ldots v_{i}^2 \geq \frac{K^2}{i}\]\[x_1^2 + x_2^2 + \ldots x_{2019 - i}^2 \geq \frac{K^2}{2019 - i}\]and therefore\[(v_1^2 + \ldots + v_{i}^2) + (x_1^2 + \ldots + x_{2019 - i}^2) = 1 \leq \frac{2019K^2}{i \cdot (2019 - i)} \leq \frac{2019 \cdot iv_1 \cdot (2019 - i)x_1}{i(2019 - i)} = 2019v_1x_1\]implying our desired result. $\blacksquare$

not a new solution but wtvr Rephrase the problem as follows: we have $a_1 \leq a_2 \cdots \leq a_n$, and $b_1 \leq \cdots \leq b_p$. Additionally, we have $$S = a_1 + a_2 + \cdots + a_n = b_1 + \cdots + b_p,$$and $$a_1^2 + \cdots + a_n^2 + b_1^2 + \cdots + b_p^2 = 1.$$Also $n + p = 2019$. Here, $n$ is for negatives, and $p$ is for positives, though it really doesn't matter. Now, notice that $S \leq na_n$ and $S \leq pb_p$ by assumption. We have the following inequalities: $$ \begin{aligned} \frac{1}{2019} &= \frac{a_1^2 + \cdots + a_n^2 + b_1^2 + \cdots + b_p^2}{2019} \\ &\leq \frac{a_nS + b_pS}{2019}. \end{aligned}$$ To finish, we need one final bound on $a_nS + b_pS$. By using our $S \leq na_n$ and $S \leq pb_p$ from earlier, we get $$Sb_p + Sa_n \leq na_nb_p + pa_nb_p = 2019 a_nb_p,$$so combining, we find $\frac{a_nS + b_pS}{2019} \leq a_nb_p$, as desired.

Notice that $(a-u_i)(b-u_i) \le 0$ for all $1 \le i \le 2019$. Summing over all $i$, we have \[ 0 \ge \sum_{i=1}^{2019} (a-u_i)(b-u_i) = \sum_{i=1}^{2019} (ab-u_i(a+b)+u_i^2) = \bigg( \sum_{i=1}^{2019} ab \bigg) - (a+b)\bigg( \sum_{i=1}^{2019} u_i \bigg) + \bigg( \sum_{i=1}^{2019} u_i^2 \bigg) = 2019ab+1, \]so $ab \le -\frac{1}{2019}$, as desired.

We have a more generalized case. Assume WLOG, $u_1 \geq u_2 \geq \cdots \geq u_n$ with $u_{1}+u_{2}+\cdots+u_{n}=0$. We will show $$-nu_1u_n \geq u_1 ^2+ u_2 ^2+ \cdots + u_n ^2$$Consider $$\begin{array}{l} \left(u_1-u_2\right)\left(u_2-u_n\right)= u_1u_2-u_1u_n-u_2^2+ u_2u_n \geq 0 \\ \left(u_1-u_3\right)\left(u_3-u_n\right)= u_1u_3-u_1u_n-u_3^2+ u_3u_n \geq 0 \\ \vdots \\ \left(u_1-u_{n-1}\right)\left(u_{n-1}-u_n\right) = u_1u_{n-1}-u_1u_n-u_{n-1}^2+ u_nu_{n-1} \geq 0 \end{array}$$Summing up, we get $$\begin{aligned} & u_1\left(u_2+\ldots+u_{n-1}\right)-n u_1 u_n-\left(u_2^2+\cdots+u_{n-1}^2\right) \\ & +u_n\left(u_2+u_3+\ldots+u_{n-1}\right) \geq 0 \\ \Rightarrow & -u_1\left(u_1+u_n\right)-n u_1 u_n-\left(u_2^2+\ldots+u_{n-1}^2\right) \\ & -u_n\left(u_1+u_n\right) \geq 0 \\ \Rightarrow & -(n+2) u_1 u_n \geq u_1^2+u_2^2+\ldots+u_n^2 \end{aligned}$$Hence we have $$ -nu_1u_n \geq -(n+2)u_1u_n \geq u_1 ^2+ u_2 ^2+ \cdots + u_n ^2$$Now put $n=2019$ and $u_1 ^2+ u_2 ^2+ \cdots + u_n ^2=1$ to get our desired result

Let $u_{2019} \ge u_{2018} \ge ... \ge u_{x+1} \ge 0 > u_x \ge u_{x-1}... \ge u_1$ Then $$xu_1u_{2019} \le u_{2019}(u_x+...+u_1)=-u_{2019}(u_{x+1}+...+u_{2019}) \le -(u_{x+1}^2+...+u_{2019}^2) \qquad (1)$$$$(2019-x)u_1u_{2019} \le u_{1}(u_{x+1}+...+u_{2019})=-u_{1}(u_{1}+...+u_{x}) \le -(u_{1}^2+...+u_{x}^2) \qquad (2)$$Adding $1$ and $2$ implies $$xu_1u_{2019}+(2019-x)u_1u_{2019} \le -(u_1^2+...+u_{2019}^2) \implies u_1u_{2019} \le -\frac{1}{2019} \qquad \blacksquare$$

$\color{green} \boxed{\textbf{SOLUTION}}$ Let, $$u_1 \le u_2 \le ... u_k$$$$u_{k+1} \ge u_{k+2} \geq ... u_{k+l}$$ $$u_i=-a_i, i \in (1,k),u_j=b_j, j \in (k+1,k+l), k+l=2019$$ So, $$a_1 \ge a_2 \ge \dots \ge a_k$$and $$b_{k+l} \le b_{k+2} \le \dots \le b_{k+1}$$ Then we have, $$\sum_{i=1}^{k} u_i + \sum_{j=k+1}^{k+l} u_j= -\sum_{i=1}^{k} a_i + \sum_{j=k+1}^{k+l} b_j =0 \implies \sum_{i=1}^{k} a_i = \sum_{j=k+1}^{k+l} b_j$$ And, $$\sum_{i=1}^{k} {a_i}^2 + \sum_{j=k+1}^{k+l} {b_j}^2=1$$ Now, $$\sum_{i=1}^{k} {a_i}^2 \leq a_1a_1+a_1a_2+...a_1a_k \le a_1b_{k+1}+a_1b_{k+2}+...a_1b_{k+l} \le la_1b_{k+1}$$ Similarly, $$\sum_{i=1}^{k} {b_j}^2 \le b_{k+1}b_{k+1}+b_{k+1}b_{k+2}+...b_{k+1}1b_{k+l} \le b_{k+1}a_1+b_{k+1}a_2+...b_{k+1}a_k \le ka_1b_{k+1}$$ Summing these, $$1=\sum_{i=1}^{k} {a_i}^2 + \sum_{j=k+1}^{k+l} {b_j}^2 \le (k+l)a_1b_{k+1} \implies (-a_1)b_{k+1} \le -\frac{1}{2019} \implies u_1u_{k+1} \le -\frac{1}{2019} \blacksquare$$

I love being able to CONVERT problems with a condition on sum $x_i$ and $x_i^2$ into the COMPUTATION of the variance of a RANDOM VARIABLE Let $X$ be a random variable uniformly distributed among the multiset $\{u_1,\ldots,u_{2019}\}$. Then $\mathbb{E}[X]=0$ and $\mathbb{E}[X^2]=\tfrac{1}{2019}$, so $\mathrm{Var}(X)=\tfrac{1}{2019}$. Thus we will prove more generally that given a random variable $Y$ sampled (with positive probability) from a finite set $S$ with $\mathbb{E}[Y]=0$, we have $\min(Y)\max(Y)\leq -\mathrm{Var}(Y)$. (This should actually be true for all $Y$ but we're not going to prove it). By scaling it obviously suffices to fix $\max(Y)=M$ and $\min(Y)=-m$ for $M,m>0$, and then show that $\mathrm{Var}(Y) \leq Mm$. Suppose that $S$ contains some element $x$ other than $M$ and $-m$. Then we can delete it and "redistribute" its probability to $M$ and $-m$, defining a random variable $Y'$ such that $\mathbb{P}(Y'=x)=0$, $\mathbb{P}(Y'=r)=\mathbb{P}(Y=r)$ for all $r \in S \setminus \{x,M,-m\}$, and $\mathbb{E}[Y']=0$. It's obvious that $\mathbb{P}(Y'=M)>\mathbb{P}(Y=M)$ and $\mathbb{P}(Y'=-m)>\mathbb{P}(Y=-m)$, so by Jensen we have $\mathrm{Var}(Y')>\mathrm{Var}(Y)$ (alternatively you can just write everything out explicitly). Doing this to all $x \in S \setminus \{M,-m\}$ (one at a time) ends up giving us the unique random variable $Z$ sampled from $\{M,-m\}$ with expected value $0$. This means that $Z$ maximizes the variance of any random variable sampled from a finite set with expectation $0$. We can now explicitly compute the variance of $Z$ as $Mm$, so we're just done. $\blacksquare$

WLOG let $u_1 \ge u_2 \ge \dots \ge u_k \ge 0 \ge u_{k+1} \ge \dots \ge u_{2019}$. Now note that $u_1 + \dots + u_k = u_{k+1} + \dots + u_{2019}$ and $u_1^2 + u_2^2 + \dots + u^2_{2019} = 1$. We know that $$\frac1{2019} = \frac{u_1^2 + u_2^2 + \dots u_{2019}^2}{2019} \le \frac{ku_1^2 + (2019 - k)u_{2019}^2}{2019} \le 2\sqrt{k(2019-k)} \left( \frac{-u_1u_{2019}}{2019} \right) \le - u_1u_{2019} $$which is equivalent to the desired result.

Say without loss of generality $u_1 \leq u_2 \leq \dots \leq u_{2019}$. Also lets say that for all $i \leq a$ we have $u_i < 0$ and for all $i > 2019 - b$ we have $u_i > 0$. It is clear to minimize the magnitude of $u_1u_{2019}$ we require all positive $u_i$ to be equal and all negative $u_i$ to be equal. Thus we have something of the form, \begin{align*} (\underbrace{-k/a, -k/a, \dots, -k/a}_{a \text{ terms}}, 0, 0, \dots, 0, \underbrace{k/b, k/b, \dots, k/b}_{b \text{ terms}}) \end{align*}for some positive $k$. Then from the second condition of squares summing to $1$ we find, \begin{align*} \frac{k^2}{a} + \frac{k^2}{b} = 1 \end{align*}Note that, \begin{align*} \frac{k^2}{ab} &= \frac{1}{a+b}\\ \iff \frac{k^2}{ab} &\geq \frac{1}{2019}\\ \iff -\frac{k^2}{ab} &\leq -\frac{1}{2019} \end{align*}This bound follows as there are $a$ positive terms and $b$ negative terms. Thus there are $a + b \leq 2019$ non-negative $u_i$. Note however that the minimum times the maximum is exactly, \begin{align*} \left(- \frac{k}{a} \right) \cdot \left( \frac{k}{b} \right) = -\frac{k^2}{ab} \end{align*}and so we are actually done.

I proudly present the most annoying and wordy solution. Replace $2019$ with $N$. Let $n,p$ denote the positive and negative terms. We know, $|\text{sum of positive terms}|=|\text{sum of negative terms}|$. Note, $\text{sum of positive terms}>pa$ and $\text{sum of negative terms}<nb$. Also, $\text{sum of squares of positive terms}\leq |b|\cdot\text{sum of positive terms}\leq |ab|p$ while $\text{sum of squares of negative terms}\leq |a|\cdot|\text{sum of negative terms}|\leq |ab|n$. Adding, we get the result thar $|ab|(n+p)=|ab|N\geq1$ as desired.

Standard equality problem. Replace $2019$ with $n.$ Note that for all $i,$ we have $(u_i - a)(u_i - b) \le 0.$ Summing over all $i$ gives \[ u_1^2 + u_2^2 + \dots + u_n^2 - (a+b)(u_1 + u_2 + \dots + u_n) + nab \le 0. \]Plugging in the given identities for the $u_i$ gives \[ 1 - 0 + nab \le 0 \rightarrow ab \le -\frac{1}{n}, \]as desired. Equality holds when $u_1 = -\sqrt{\frac{n-1}{n}}$ and $u_i = \frac{1}{\sqrt{n(n-1)}}$ for $i \ge 2.$

Let $n=2018$. WLOG assume $u_{n+1}=u \geq u_1 \geq u_2 \geq... \geq u_n$. Define $u_i=u-d_i$ there $i=1, 2,...,n$, so $d_n\geq d_{n-1}\geq...\geq d_1 \geq 0$. By condition we have $(n+1)u=d_1+... +d_n$ and $1=(n+1)u^2-2u(d_1+... +d_n)+(d_1^2+... +d_n^2)=(d_1^2+... +d_n^2)-u^2(n+1)$. By this reason $u^2(n+1)+1\leq d_n(d_1+... +d_n)=d_n(n+1)u$, i.e. $(n+1)ab+1=(n+1)u(u-d_n)+1=(n+1)u^2+1-d_n(n+1)u \leq 0$ and we are done!

Why did I try literally every local estimate possible except for this one :/ Sum cyclically the estimate $(u_i-a)(u_i-b) \leq 0$ to get \[u_1^2 + u_2^2 + \cdots + u_{2019}^2 - (a+b)(u_1+u_2+\cdots+u_{2019}) + 2019ab \leq 0.\]Plugging in the given values yields $ab \leq -\frac 1{2019}$, as needed.

Consider the sum \[\sum_i (u_i - a)(u_i - b)\]and clearly each term is nonpositive so by expanding we get \[(\sum_i u_i^2 + ab) \leq 0 \implies ab \leq -\frac{1}{2019}\]as desired.

i've been spoiled to the solution but it feels strangely motivated, like at least the first thing that comes to mind when you see the bounds but then again i got spoiled so i was probably disposed to thinking the correct solution was obvious idk oh well. Notice that \[(u_i-a)(u_i-b)\le 0\implies ab-u_i(a+b)+u_i^2\le 0\]for each $1\le i\le 2019$. Summing over all $i$ yields: \[2019ab-(u_1+\dots+u_{2019})(a+b)+(u_1^2+\dots+u_{2019}^2)\le 0\implies 2019ab+1\le 0\implies ab\le -\frac{1}{2019}\]and we are done. $\blacksquare$

The condition $u_1^2+u_2^2+...+u_{n}^2=1$ is useless, in a such way that, if $\sum_i u_i^2=\mathbb{S}$, we can prove that $$ab \leq -\frac{\mathbb{S}}n$$So, let´s do this: First, supose WLOG $u_1=a$ and $u_n=b$ and let $X$ be a discrete random varible wich takes the value $u_1, u_2, ..., u_n$ uniformly and randomly. Since $\sum_i u_i=0$ we get that$$\mathbb{E}[X]=\sum_i (\frac{1}n \cdot u_i)=0$$Therefore$$\sigma^2_X=\mathbb{E}[X^2] -\mathbb{E}[X]^2=\sum_i (\frac{u_i^2}n)-0=\frac{\mathbb{S}}n$$Now, proceed with the following algorithim: $$\text{1. Let } X_1 \text{ be a new randomly choosen variable such that } \mathbb{P}(X_1=u_{2})=0 \text{ redistributing this probability for } a \text{ and } b \text{ in such way that } \mathbb{E}[X_1]=0;$$$$\text{2. Take } X_i \text{ and, from it, create a new random variable } X_{i+1} \text{ such that } \mathbb{P}(X_{i+1}=u_{i+2})=0 \text{ and } \mathbb{E}[X_{i+1}]=0 \text{ again redistributing this probability for } a \text{ and } b;$$$$\text{3. For } 1\leq i\leq n-3 \text{ repeat 2. }$$Hence, by the end of this algorithm we´ll have $X_{n-2} \in \{a, b\}$ with $\mathbb{E}[X_{n-2}]=0$, and, as the probabilty of the value of $X_{i+1}$ is more concentrated in extreme values like $a$ and $b$ than $X_i$ and $\mathbb{E}[X_{i+1}]=\mathbb{E}[X_i]=0$ whe get that $$\sigma^2_{X_{n-2}} \geq \sigma^2_{X_{n-1}} \geq ... \geq \sigma^2_{X_1} \geq \sigma^2_X=\frac{\mathbb{S}}2$$But, we can explicity calculate $\mathbb{E}[X_{n-2}]$: let $\mathbb{P} (X_{n-2}=a)=p$ and $\mathbb{P} (X_{n-2}=b)=1-p$, so$$\mathbb{E}[X_{n-2}]=0=pa+(1-p)b \Rightarrow p=\frac{-b}{a-b} \Rightarrow 1-p=\frac{a}{a-b}$$$$\sigma^2_{X_{n-2}}=\mathbb{E}[X^2_{n-2}]=pa^2+(1-p)b^2=\frac{-a^2b +ab^2}{a-b}=-ab \geq \sigma^2_X=\frac{\mathbb{S}}n$$So we´re done $\blacksquare$

Notice that, $$(b-u_i)(u_i-a) \ge 0 \Rightarrow (a+b)u_i-u^2_i \ge ab$$Summing the inequalities for $i=1,2,\cdots,2019$ yields the desired result.