VicKmath7 wrote: Given are real numbers $a_1, a_2,...,a_{101}$ from the interval $[-2,10]$. Prove that the sum of their squares is smaller than $2020$. Going by your question, I think you meant $distinct$ real numbers, or else the maximum sum possible will be $10^2*101=10100$

Corrected now, their sum is 0.

$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Proof}$ For all $a\in\mathbb{R}$ holds $$a\in[-2,10]\iff (a+2)(a-10)\le 0\iff a^2\le 8a+20.$$Therefore $$\sum_{i=1}^{101}a_i^2\le\sum_{i=1}^{101}(8a_i+20)=20\cdot 101=2020.$$Can there be equality? It holds iff $k$ numbers is equal to $-2$ and $101-k$ numbers is equal to $10$. It comes to $$0=\sum_{i=1}^{101}a_i=-2k+10(101-k)=10\cdot 101-12k\implies k=\frac{10\cdot 101}{12}\notin\mathbb{Z}.$$Hence no equality and finally $$\definecolor{A}{RGB}{240,60,60}\color{A}\sum_{i=1}^{101}a_i^2<2020.\blacksquare$$#1730

nice solution! use the same method,we can prove that: for $a_1,a_2,\cdots,a_n \in [A,B]$, and $a_1+a_2+\cdots+a_n=0$, we have $a_1^2+a_2^2+\cdots+a_n^2 \leq -nAB$.

Notice that $(x-10)(x+2) \leq 0$ for $x \in [-2,10]$. Thus... $(a_1-10)(a_1+2)+ \cdots (a_{101}-10)(a_{101}+2) \leq 0$ $\Rightarrow (a_1^2+ \cdots +a_{101}^2)-8(a_1+ \cdots +a_{101}) \leq 2020$ $\Rightarrow a_1^2+ \cdots +a_{101}^2 \leq 2020$ Equality holds if $a_i=-2$ or $10$ for all $i$. If there are $n$ $-2$'s and $101-n$ $10$'s then. $a_1^2+ \cdots +a_{101}^2=4n+100(101-n)=10100-96n$. But $2020 \not \equiv 10100$ (mod $96$). Contradiction.

easy-proof wrote:

nice solution! use the same method,we can prove that: for $a_1,a_2,\cdots,a_n \in [A,B]$, and $a_1+a_2+\cdots+a_n=0$, we have $a_1^2+a_2^2+\cdots+a_n^2 \leq -nAB$. Yes, this is exactly Problem A2 from the IMO Shortlist 2019 (which actually, turns out, already appeared on the IMO Shortlist 1972).

You can also use the equal-values-principle (which results in a less elegant solution):

Notice that \[(a_i + 2)(a_i + 10) \leq 0\]$\implies$ \[a_i(a_i - 8) \leq 20\]Which is always true, and equality is achieved when $a_i = 10$, or $a_i = -2$. Summing this $101$ times for each $a_i$ results in $2020$, but since the sum of all $a_i$ is $0$, we cannot have equality, because not all $a_i$ can equal $10$ or $-2$, because there are $101$ terms.

We have, \begin{align*} (a_i + 2)(10 - a_i) &\geq 0\\ 8a_i + 20 &\geq a_i^2 \end{align*}Then summing we have, \begin{align*} 2020 &\geq \sum a_i^2 \end{align*}We cannot have equality as there are $101$ terms.

Our domain restriction for $\{a_n\}$ tells us \[0 \ge \sum (a_i+2)(a_i-10) = \left(\sum a_i^2\right) - 2020.\] Note that equality only holds when each $a_i \in \{-2, 10\}$, which isn't possible. $\blacksquare$

We have $(a_i + 2)(a_i - 10) \le 0$ for all $1 \le i \le 101$. Summing among all $i$ yields the conclusion. (The inequality is strict as we cannot have all variables either $-2$ or $10$)

\textit{Solution 1:} For numbers $a,b$ such that $a<b$, we see that if $\epsilon$ is positive, $$(a-\epsilon)^2 + (b+\epsilon)^2 = a^2 + b^2 + 2 \epsilon(b-a) + 2 \epsilon^2 \ge a^2+b^2.$$Thus by smoothing (more like anti-smoothing) argument, by making the numbers farther apart, the sum of squares becomes greater. This operation cannot be applied further if all (or all except one) elements are at the limits ($-2, 10$). Thus, let our set have $16$ of $10$'s, $84$ of $-2$'s, and one $8,$ which gives a sum of square value of $2000.$ \qed \textit{Solution 2:} Because of the bounds, we have $(a_i+2)(a_i-10)\le 0.$ Expanding gives $$a_i^2 \le 8a_i + 20,$$so $$\sum a_i^2 \le 8\sum a_i + \sum 20 = 20 \cdot 101 = 2020.$$After some experimentation, the equality case $a_i \in \{ -2, 10 \}$ cannot be realized, so the inequality is strict. \qed