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Source: Serbia JBMO TST 2020 P3

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1389 posts

#1 h Sep 5, 2020, 6:22 AM • 2 Y

Y by itslumi, ItsBesi

Given are real numbers $a_1, a_2,...,a_{101}$ from the interval $[-2,10]$ such that their sum is $0$ . Prove that the sum of their squares is smaller than $2020$ .

This post has been edited 3 times. Last edited by VicKmath7, Sep 5, 2020, 6:45 AM

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#2 h Sep 5, 2020, 6:27 AM • 1 Y

Y by Mango247

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 $10^2*101=10100$

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1389 posts

#3 h Sep 5, 2020, 6:32 AM • 3 Y

Y by Mango247, Mango247, Mango247

Corrected now, their sum is 0.

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1900 posts

#4 h Sep 5, 2020, 7:15 AM • 5 Y

Y by peaceGiant, lahmacun, domce, Math_legendno12, kiyoras_2001

$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Proof}$
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

This post has been edited 4 times. Last edited by WolfusA, Sep 5, 2020, 12:00 PM

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#5 h Sep 5, 2020, 8:58 AM

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solution

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 $a_1^2+a_2^2+\cdots+a_n^2 \leq -nAB$ .

This post has been edited 3 times. Last edited by easy-proof, Sep 5, 2020, 10:48 AM

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#6 h Jul 4, 2021, 7:01 PM

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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.

This post has been edited 2 times. Last edited by blackbluecar, Aug 7, 2021, 11:58 PM

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#7 h Oct 23, 2022, 7:57 PM

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solution

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9042 posts

#8 h Oct 24, 2022, 8:42 AM

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easy-proof wrote:

solution

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).

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#9 h Oct 24, 2022, 12:05 PM

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

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Taco12

#10 h Mar 3, 2023, 11:59 PM

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Sketch

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#11 h Jan 5, 2024, 9:33 PM

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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.

This post has been edited 2 times. Last edited by dolphinday, Jan 5, 2024, 9:35 PM

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#12 h Feb 15, 2024, 5:24 AM

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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.

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#13 h Feb 18, 2024, 2:33 AM

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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. $\blacksquare$

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#14 h Aug 5, 2024, 8:10 PM

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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 $10$ )

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bebebe

#15 h Aug 22, 2024, 5:57 PM

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\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

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eg4334

#16 h Jan 31, 2025, 3:42 AM • 1 Y

Y by Marcus_Zhang

Each term satisfies $(a+2)(a-10) \leq 0$ . Summing gives the desired conclusion because $20 \cdot 101 = 2020$ . To see that equality cannot work, assume all are $-2$ or $10$ . Then $10x-2(101-x)=0$ which does not have an integer solution.

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#17 h Apr 7, 2025, 2:46 AM

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This took me way too long

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