For 1 ≤ i ≤ 215 let a i = 1 2 i and a 216 = 1 2 215 . Let x 1 , x 2 , … , x 216 be positive real numbers such that ∑ i = 1 216 x i = 1 and ∑ 1 ≤ i &lt; j ≤ 216 x i x j = 107 215 + ∑ i = 1 216 a i x i 2 2 ( 1 − a i ) .Find the maximum possible value of x 2 .I simplified the condition to ∑ i = 1 216 x i 2 1 − a i = 1 215 , but I'm not sure what to do next.

Question

For   1  ≤  i  ≤  215 let       a    i    =            1              2                  i                     and       a          216        =            1              2                  215                    . Let       x    1    ,      x    2    ,  …  ,      x          216       be positive real numbers such that             ∑              i        =        1                    216                    x      i        =    1   and      ∑          1      ≤      i      &amp;lt;      j      ≤      216            x    i        x    j    =            107      215        +      ∑          i      =      1              216                                    a          i                          x          i                      2                                      2        (        1        −                  a          i                )              .Find the maximum possible value of       x    2    .I simplified the condition to             ∑              i        =        1                    216                                      x          i          2                          1          −                      a            i                                =                  1        215              ,   but I&#039;m not sure what to do next.

livin4him777lf · Accepted Answer

Given   ∑            x      i      2              1      −              a        i              =      1    215  , apply Cauchy Schwarz / Titu&#039;s lemma to get  ∑            x      i      2              1      −              a        i              ≥            (      ∑              x        i                    )        2                    ∑      1      −              a        i              =            1      2        215    .Since equality holds throughout, we conclude that             x      i              1      −              a        i             is a constant, say   k.Since   1  =  ∑      x    i    =  ∑  (  1  −      a    i    )  k  =  215  k, so   k  =      1    215  .Hence,       x    i    =            1      −              a        i              215  .

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