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Let $a_1,a_2,a_3,a_4,a_5$ be real numbers such that $a_1 + a_2 + a_3 + a_4 + a_5=0$ and $\displaystyle\max_{1\le i<j\le 5} |a_i - a_j|\le1$. Prove that $a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 \le 10$.

I am not able to understand what $\displaystyle\max_{1\le i<j\le 5} |a_i - a_j|\le1$ means or how to extract useful information from this expression, maybe because I am not mathematically mature enough. How should I approach this problem?

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  • $\begingroup$ The statement means that for all 5 of your elements $a_i$, the distance between any of them is not higher than $1$. Meaning that for any two elements you take, their difference must be a real number between -1 and 1 $\endgroup$ – Francisco José Letterio May 30 '18 at 12:26
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The condition gives $$\sum_{1\leq i<j\leq5}(a_i-a_j)^2\leq10$$ or $$4\sum_{i=1}^5a_i^2-2\sum_{1\leq i<j\leq5}a_ia_j\leq10$$ or $$5\sum_{i=1}^5a_i^2-\left(\sum_{i=1}^5a_i\right)^2\leq10,$$ which gives $$\sum_{i=1}^5a_i^2\leq2<10.$$ Because since $(a-b)^2=a^2-2ab+b^2,$ we obtain $$\sum_{1\leq i<j\leq5}(a_i-a_j)^2=(a_1-a_2)^2+(a_1-a_3)^2+(a_1-a_4)^2+(a_1-a_5)^2+$$ $$+(a_2-a_3)^2+(a_2-a_4)^2+(a_2-a_5)^2+(a_3-a_4)^2+(a_3-a_5)^2+(a_4-a_5)^2=$$ $$=4(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)-$$ $$-2(a_1a_2+a_1a_3+a_1a_4+a_1a_5+a_2a_3+a_2a_4+a_2a_5+a_3a_4+a_3a_5+a_4a_5)=$$ $$=5(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)-(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+$$ $$+2(a_1a_2+a_1a_3+a_1a_4+a_1a_5+a_2a_3+a_2a_4+a_2a_5+a_3a_4+a_3a_5+a_4a_5))=$$ $$=5(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)-(a_1+a_2+a_3+a_4+a_5)^2=5(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2).$$

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  • $\begingroup$ What was your thought process, how did you approach this problem? $\endgroup$ – user521346 May 30 '18 at 10:50
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    $\begingroup$ @Simba Any $(a_i-a_j)^2\leq1$ and the number of numbers $(a_i-a_j)^2$ is $\binom{5}{2}=10$. Thus, the sum of them is $\leq10$. $\endgroup$ – Michael Rozenberg May 30 '18 at 10:53
  • $\begingroup$ what is the difference between $\max_{1\le i<j\le 5} |a_i - a_j|\le1$ and simply $|a_i - a_j|\le1$ for $1\le i<j\le 5$ what role does $\max$ play? $\endgroup$ – user521346 May 30 '18 at 11:05
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    $\begingroup$ @Simba Just $\max\limits_{1\leq i<j\leq5}|a_i-a_j|\leq1$ says that all $|a_i-a_j|\leq1$. For example, $\max\{0.1,1\}\leq1$ says $0.1\leq1$ and $1\leq1$. $\endgroup$ – Michael Rozenberg May 30 '18 at 11:22
  • $\begingroup$ So it is same as $|a_i-a_j|\leq1$ $\forall 1\le i<j\le 5$ $\endgroup$ – user521346 May 30 '18 at 11:28
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What the question means is that absolute value of the difference between any two of those five numbers is less than or equal to $1$.

Since the sum of those five numbers is equal to zero, this should necessarily mean that all five numbers are in the interval $(-1, 1)$.

Hence the value of the sum of their square can never be more than $5$, which solves the problem.

Note - I believe that you may have mis-typed the question since the bound is too weak!

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  • $\begingroup$ I think the sum of their squares can never be more than 2, don't you think? (Instead of 5) $\endgroup$ – user521346 May 30 '18 at 10:48
  • $\begingroup$ @Simba I am not mathematically familiar enough with inequalities to show that the bound is $2$, but when I saw the question has a bound of $10$ a simple proof came to my mind which proves that the bound is _atleast equal to or less than $5$_(and hence solves the particular) I hope that it is correct and (at least somewhat) useful:) I am very interested to see Wen's proof of the bound being $1.2$. $\endgroup$ – Agile_Eagle May 30 '18 at 10:50
  • $\begingroup$ what is the difference between $\max_{1\le i<j\le 5} |a_i - a_j|\le1$ and simply $|a_i - a_j|\le1$ for $1\le i<j\le 5$ what role does $\max$ play? $\endgroup$ – user521346 May 30 '18 at 11:06
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    $\begingroup$ @Simba I think both statements mean the same, they are different ways of writing (notation) the same thing. $\endgroup$ – Agile_Eagle May 30 '18 at 11:07
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Because of symmetry we can assume that $x:=a_1\leq a_2\leq a_3\leq a_3\leq a_5$.

So $a_i-a_1 \leq 1 \implies a_i\leq x+1$ for each $i=2,3,4,5$.

Since $$5x\leq \underbrace{a_1+a_2+...a_5}_{=0}\leq 5x+4$$ we get $-{4\over 5}\leq x\leq 0$

Since $x\in[-{4\over 5},0]$ we have $a_2,a_3,a_4,a_5\leq 1$ so

Now we have $$a_1^2+...+a_5^2\leq 4+{16\over 25}<10$$

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$\max_{1\le i<j\le 5} |a_i - a_j|\le1$

means

$|a_x - a_y|\leq\max_{1\le i<j\le 5}|a_i - a_j|\le1$

for any $x,y$ (by definition of the max function)

See if you can solve it from there

Hint:

Since one of the $a_i$s is not positive, and one is not negative (otherwise they can't sum to $0$), what is the largest any $a_i$ can be? What about the smallest value?

(Edit I think $a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 \le 1.2$ is true as well - is that what you meant?)

I have a proof of the $\le1.2$ if you want it

Consider $$\sum_{i<j}(a_i-a_j)^2=\sum_{i<j}(a_i-a_j)^2+(\sum a_i)^2=5\sum a_i^2$$

Now assume $a_1\leq a_2\leq\ldots\leq a_5$. Note that $(c-a)^2\geq(c-b)^2+(b-a)^2$ if $c\geq b\geq a$ (think about this geometrically if that helps). Applying this to the LHS gives us $$\begin{align} 6\geq &(a_5-a_4)^2+(a_4-a_3)^2+(a_3-a_2)^2+(a_2-a_1)^2+\\ &(a_5-a_4)^2+(a_4-a_2)^2+(a_2-a_1)^2+\\ &(a_5-a_4)^2+(a_4-a_1)^2+\\ &(a_5-a_3)^2+(a_3-a_1)^2+\\ &(a_5-a_2)^2+(a_2-a_1)^2+\\ &(a_5-a_1)^2\geq\\ &\sum_{i<j}(a_i-a_j)^2 \end{align}$$

since each line is at most $1$. Therefore $6\geq5\sum a_i^2$ as required (also this is attainable, e.g $(0.4,0.4,0.4,-0.6,-0.6)$)

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  • $\begingroup$ Yes I want (random letters to satisfy "at least 15 characters in length") $\endgroup$ – user521346 May 30 '18 at 10:45
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The following is a proof for a bound of 1.2: $$$$ Without loss of generality, let $$a_1\le a_2 \le ... \le a_5$$ let $$(i,j)= |a_i - a_j|$$ Now we have $$\sum_{i=1}^{5}{(i,i+1)} = \sum_{i=1}^{4}{(i,i+1)} + (5,1) = 2 \times (1,5) < 2 $$ where i is taken mod 5. Also, $$\sum_{i=1}^{5}{(i,i+2)} = (1,3)+(2,4)+(3,5)+(1,4)+(2,5) \le 4(1,5) \le 4 $$ (this is clear from drawing the 5 numbers on a number line and drawing segments. So now we have $$\sum_{1\le i < j \le 5} (i,j) \le 6$$ Note that $$0\le(i,j)\le 1 \implies (i,j)^2 \le (i,j),$$ so $$\sum_{1\le i < j \le 5} (i,j)^2 \le 6 \implies 4\sum_{i=1}^{5}{a_i}^2-2\sum_{1\le i < j \le 5}{a_i a_j} \le 6$$ By squaring the first condition, $$\sum_{i=1}^{5}{a_i}^2 = -2\sum_{1\le i < j \le 5}{a_i a_j}$$ So substituting this, we find that $$5\sum_{i=1}^{5}{a_i}^2\le 6$$ which gives the desired bound of 1.2. $$$$Careful analysis of the solution shows that $${0.4, 0.4, 0.4, -0.6, -0.6}$$ and $${0.2, 0.2, 0.2, 0.2, -0.8}$$ as well as negation of each term and permutations are the only equality cases.

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