1400 real numbers are given. Prove that one can choose three of them like x,y,z such that : |(x−y)(y−z)(z−x)x4+y4+z4+1|<0.009
Problem
Source: 2021 Iran second round mathematical Olympiad P5
Tags: combinatorics, algebra, inequalities, Iran
16.05.2021 11:31
Yaghi wrote: 1400 real numbers are given. Prove that one can choose three of them like x,y,z such that : |(x−y)(y−z)(z−x)x4+y4+z4+1|<0.009 Just choose thrice the same number ... .
16.05.2021 14:04
Quote: Just choose thrice the same number ... . Well actually x,y,z are pairwise distinct. However Yaghi thinks saying "three of them" implies that they are distinct! Even after me and Saeed Agha Haghi had an intense argument with him to convince him that it doesn't, he refused to edit the post; so I decided to correct it in a comment.
16.05.2021 14:06
the numbers should be different (anyways it wasn't mentioned in the actual exam )
16.05.2021 16:37
pco wrote: Yaghi wrote: 1400 real numbers are given. Prove that one can choose three of them like x,y,z such that : |(x−y)(y−z)(z−x)x4+y4+z4+1|<0.009 Just choose thrice the same number ... . you can't choose a number more than once ( not stated on exam paper , but without this assumption , problem would be obvious )
16.05.2021 19:16
Since dealing with negative numbers is annoying, without loss of generality, assume there are at least 700 (513 is enough) positive numbers among the given 1400 real numbers 0<a1<⋯<a700=bDivide the interval [0,b] into 28 sub-intervals Ik=[k28b,p+k+128b],0≤k≤28−1. By the pingeonhole principle, there are 3 numbers x>y>z belonging to the same interval Ik. For these 3 numbers x,y,z we have |(x−y)(y−z)(z−x)x4+y4+z4+1|≤(x−y)(y−z)(x−z)≤14(x−z)3≤14(b28)3=b3226If b≤64 then the desired quantity does not exceed 643226=0.0390625 and we are done. Otherwise b>64 and by choosing x=a700=b and y>z arbitrary we have |(x−y)(y−z)(z−x)x4+y4+z4+1|=(b−y)(y−z)(b−z)b4+y4+z4+1≤14b3b4<14b<14×64=0.00390625 PS: I think we can get much tighter bounds, anyone has any ideas?
18.05.2021 20:43
A logical way... If we have two equal numbers, the statement will be obvious. On the other hand, we can assume that at least 700 of them are positive. Suppose x>y>z≥0 Now the equation will become: (x−y)(y−z)(x−z)x4+y4+z4+1<0.009⟺9(x4+y4+z4+1)>1000(x−y)(y−z)(x−z)=1000(x2y−x2z−xy2+xz2−yz2+y2z)⟺9(x4+y4+z4+1)+1000(x2z+xy2+yz2)>1000(x2y+xz2+y2z)We know that: xy2>xz2,x2z>y2zSo it is enough to prove: 9(x4+y4+z4+1)+1000yz2>1000x2y⟺x2(9x2−1000y)+z2(9z2+1000y)+9y4+9>0If ∃x such x≥10009→9x2>1000y and the inequality will be obvious. Now consider all 700 numbers are less than 10009 : a1<a2<a3<.....<a700<10009Let c=a2k+1−a2k−1 be the min between following 349 expressions: a3−a1,a5−a3,…,a699−a697Consider x>y>z as a2k+1>a2k>a2k−1 we'll have: x−z≤c,(y−z)(x−y)≤c24AM−GM→(x−y)(y−z)(x−z)x4+y4+z4+1<c34Now thats enough to have c34≤91000⟺c≤3√3610 For the sake of contradiction, let us assume: c>3√3610: →a699=a1+(a3−a1)+(a5−a3)+.....+(a699−a697)≥a1+349c≥a1+3493√3610>115.23We arrive at a contradiction: 10009 111.11>a699>115.23
19.05.2021 02:22
@mhr2004 Can you obtain a better upper bound?
19.05.2021 04:59
Is there a generalisation with n- reals numbers ?
19.05.2021 06:14
@Moubinool It depends on what kind of generalizations you want. For example, using my method we can show that among 2n+1+1 positive real numbers we can choose 3 numbers x,y,z satisfying |(x−y)(y−z)(z−x)x4+y4+z4+1|<1234n+2 This bound is, in my opinion, extremely weak but unfortunately I don't know how to do better.
19.05.2021 13:37
After discussing with @mhr2004, I managed to lower the upper bound to 12000=0.0005. The most crucial idea of this proof is his. I'm reusing the notations from my post #6. Also denote f(x,y,z) the given expression for brevity. For starter, if b≥500 then by the estimate I established at the end of post #6, f(b,y,z)<14b≤14×500=12000so from now on, consider b<500. If b≥160, there are two sub-cases to consider ⋆ There are two other numbers y>z in the interval [825b,b]. In this case f(b,y,z)<14⋅(17b25)3b4+(825b)4+(825b)4=1228251595268b<1228251595268×160≈0.0004812 ⋆ At least 699 numbers belongs to the interval [0,825b) whose length is 825b<160 Thus in the case b<500, we can always suppose that at least 698 numbers a1,…,a698 belong to [0,160). Now consider 160 intervals [0,1),[1,2),…,[159,160). If there are 5 numbers in some interval [k,k+1), k≥3 then at least 3 numbers lie in either [k,k+12) or [k+12,k+1). In either cases we can bound f(x,y,z)≤14⋅(12)33k4+1≤14⋅(12)33⋅34+1≈0.000128 Otherwise in each interval [k,k+1), 3≤k≤159 there are at most 4 numbers. So there are at least 698−157×4=70 numbers distributed among the three intervals [0,1),[1,2),[2,3). Thus there is an interval containing at least 24 numbers, from which we can choose three x>y>z belonging to a sub-interval of length 18. In this case f(x,y,z)<14⋅(18)3≈0.000488so we are done!
28.05.2021 09:52
1)There are x,y,z if the following condition is met 91000≥|(x−y)(y−z)(z−x)| 91000≥|(x−y)(y−z)(z−x)|1≥|(x−y)(y−z)(z−x)|x4+y4+z4+1 2)So for every x,y,z we have that: |(x−y)(y−z)(z−x)|≥91000 a1400≥a1399≥a1398≥....≥a1 a1400−a1399,a1399−a1398,a1398−a1397,........,a2−a1 a1400−a1399,a1399−a1398,a1400−a1398,It is not possible for all three expressions to be smaller than 3√91000 a1400−a1398≥3√91000,a1398−a1396≥3√91000,a1396−a1394≥3√91000,.....,a4−a2≥3√91000 Collect all the above phrases a1400−a2≥6993√91000≥140 So one of two numbers |a1400| , |a2| bigger than 70 In a similar way, it can be concluded that |a1|,|a2|,|a3|≥69 or |a1400|,|a1399|,|a1398|≥69 1) a1400≥a1399≥a1398≥69 x≥y≥z (x−z)3x4+y4+z4+1≥|(x−y)(y−z)(z−x)|x4+y4+z4+1 Then x=a1400,y=a1399,z=a1398 if 2a1398>a1400,....... End problem otherwise a1400>138: 91000>1138>1x>(x−z)3x4+y4+z4+1 2)a1400,a1399,a1398 is negative |a1400|,|a1399|,|a1398|>69 a1400,a1399,a1398<−69 a1400−a2>140,−a2>209,a2<−209 Then Same way a1,a2,a3<−208 x=a1,y=a2,z=a3 kill problem..... And the desired result is achieved. 3)a1400,a1399,a1398 is negative |a3|,|a2|,|a1|>69........ a1,a2,a3<−69 because a1400−a2>140.....a1400>140+a2>71 then There is a contradiction. 4)a1400 is postive, a1398 is negative Very similar to previous cases
28.05.2021 13:42
x1<x2<...<x1400⟶thereexistx,y,z|(x−y)(x−z)(y−z)x4+y4+z4+1|<0.009 We claim that if three of the numbers, such as xi,xi+1,xi+2 belong to the interval (t,t+0.3)the question will be concluded.because: (xi+2−xi+1)(xi+1−xi)(xi+2−xi)x4i+x4i+1+x4i+2+1<(xi+2−xi+1)(xi+1−xi)(xi+2−xi))≤((xi+2−xi+1+xi+1−xi2)2(xi+2−xi)=14(xi+2−xi)3<(0.03)34<91000 hence Therefore, the distance between numbers is one among the at least 0.3. According to the Pigeonhole principle, at least 700 numbers are negative or positive.WLOG we suppose that: 0≤x700<x701<...<x1400Hence 0.3<x702,0.6<x704,...,699⋅0.3<x1398 ⇒x1398,x1399,x1400∈(200,∞) We prove that for these numbers is true.because (x1400−x1399)(x1399−x1398)(x1400−x1398)x41400+x41399+x41398+1<14(x1400−x1398)3x41400<14x31400x41400=14x1400<1800<1900
21.04.2022 19:49
Hopeooooo wrote: x1<x2<...<x1400⟶thereexistx,y,z|(x−y)(x−z)(y−z)x4+y4+z4+1|<0.009 We claim that if three of the numbers, such as xi,xi+1,xi+2 belong to the interval (t,t+0.3)the question will be concluded.because: (xi+2−xi+1)(xi+1−xi)(xi+2−xi)x4i+x4i+1+x4i+2+1<(xi+2−xi+1)(xi+1−xi)(xi+2−xi))≤((xi+2−xi+1+xi+1−xi2)2(xi+2−xi)=14(xi+2−xi)3<(0.03)34<91000 hence Therefore, the distance between numbers is one among the at least 0.3. According to the Pigeonhole principle, at least 700 numbers are negative or positive.WLOG we suppose that: 0≤x700<x701<...<x1400Hence 0.3<x702,0.6<x704,...,699⋅0.3<x1398 ⇒x1398,x1399,x1400∈(200,∞) We prove that for these numbers is true.because (x1400−x1399)(x1399−x1398)(x1400−x1398)x41400+x41399+x41398+1<14(x1400−x1398)3x41400<14x31400x41400=14x1400<1800<1900 this is indeed a magnificent discovery1800≤1900
22.04.2022 19:37
Indeed it is