Find the smallest possible value of ω such that the system of inequalities ( 1009 ⋅ 1010 − 2018 ) x + 2018 ≤ 2018 ω ( 1008 ⋅ 1009 − 2018 ) x + 2018 ≤ 2016 ω ( 1007 ⋅ 1008 − 2018 ) x + 2018 ≤ 2014 ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 ⋅ 2 − 2018 ) x + 2018 ≤ 2 ω has a real solution in x.Subtracting the first inequality from the second inequality, we get that x = 1009 2017 . This is also correct when we subtract the second inequality from the third one. When I plugged in x, I got an unruly fractional value for ω. Can someone guide me through the solution? Help is much appreciated.

Question

Find the smallest possible value of   ω such that the system of inequalities                    (        1009        ⋅        1010        −        2018        )        x        +        2018        ≤        2018        ω                            (        1008        ⋅              1009        −        2018        )        x        +        2018        ≤        2016        ω                            (        1007        ⋅              1008        −        2018        )        x        +        2018        ≤        2014        ω                            .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .        .                            (        1        ⋅        2        −        2018        )        x        +        2018        ≤        2        ω            has a real solution in   x.Subtracting the first inequality from the second inequality, we get that   x  =      1009    2017  . This is also correct when we subtract the second inequality from the third one. When I plugged in   x, I got an unruly fractional value for   ω. Can someone guide me through the solution? Help is much appreciated.

Paxton James · Accepted Answer

You have  (  n  (  n  +  1  )  −  2018  )  x  +  2018  ≤  2  n  wfor   n  =  1  ,  2  ,  …  ,  1009.The solution to one of these is  x      {                            ≤                                    2              n              w              −              2018                                      n              (              n              +              1              )              −              2018                                                n          ≥          45                                      ≥                                    2              n              w              −              2018                                      2018              −              n              (              n              +              1              )                                                n          ≤          44                        Where we get that   44 because the positive root of the quadratic on the denominator is  α  =            −      1      +      3              897              2    ≈  44.4You want there to exist an   x for which all of these inequalities hold. To do this, let&#039;s work out for which   n the first fraction is smallest, and for which   n the second fraction is largest.It turns out that the fractions have larger magnitude for larger   n. If   w  &amp;gt;  α  /  2 then the fraction is always negative on the range   n  ≤  44, and the maximum occurs at   n  =  1. Similarly for the other fraction, if   w  &amp;gt;  α  /  2, the minimum occurs at   n  =  1009, and if   w  &amp;gt;  α  /  2 then the minimum occurs at   n  =  45.Let&#039;s first take the case   w  &amp;lt;  α  /  2. Then we need            2      w      −      2018        2016    ≤  x  ≤            90      w      −      2018        52  As before, we equate these two sides to give   52  (  2  w  −  2018  )  =  2016  (  90  w  −  2018  ). This give   181336  w  =  3963352    ⟹        w    =    21.8564    ⋯, which satisfies   w  &amp;lt;  α  /  2, so is a valid solution. This gives the real solution of   x  =  −  0.9793091  ⋯The other case cannot produce a smaller solution, so this is the smallest possible solution.

Find the smallest possible value of ω such that the system of inequalities

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