<prompt> Kwa 1 leq i leq 215 acha a_i dfrac12i na a_216 dfrac12215. Acha x_1, x_2, dots, x_216 ziwe nambari halisi chanya kiasi kwamba sum_i1216 x_i1 nasum_1 leq i j leq 216 x_ix_j dfrac107215 sum_i1216 dfraca_i x_i22 1-a_i.Tafuta thamani ya juu iwezekanavyo ya x_2.  <response>: Kuzidisha pande zote mbili kwa 2, tunapata2x_1 x_2 2x_1 x_3 dots 2x_2015 x_2016 frac214215 sum_i 12016 fraca_i1 - a_i x_i2.Kisha ukiongeza x_12 x_22 dots x_20162, tunaweza kuandika mlinganyo kamax_1 x_2 dots x_20162 frac214215 sum_i 12016 fracx_i21 - a_i .Kwa kuwa x_1 x_2 vitone x_2016 1,1 frac214215 sum_i 1216 fracx_i21 - a_i,hivyosum_i 1216 fracx_i21 - a_i frac1215.Kutoka Cauchy-Schwarz, kushoto sum_i 1216 fracx_i21 - a_i kulia kushoto sum_i 1216 1 - a_i kulia ge kushoto sum_i 1216 x_i kulia2.Hii hurahisishafrac1215 sum_i 1216 1 - a_i ge 1,hivyojumla_i 1216 1 - a_i ge 215.Tanguanzaalignjumla_i 1216 1 - a_i 1 - a_1 1 - a_2 1 - a_3 vitone 1 - a_216 216 - a_1 a_2 a_3 vitone a_216 216 - left frac12 frac122 frac123 dots frac12 215 frac12215 kulia 216 - 1 215,endaligntuna usawa katika usawa wa Cauchy-Schwarz. Kwa hivyo, kutoka kwa hali ya usawa,fracx_i21 - a_i2ni thabiti, au sawa na fracx_i1 - a_i haibadiliki, sema c. Kisha x_i c 1 - a_i kwa i yote, hivyojumla_i 1216 x_i c jumla_i 1216 1 - a_i.Hii inatupa 1 215c, hivyo c frac 1215. Kwa hivyo,fracx_21 - a_2 frac1215,au x_2 frac1 - a_2215 frac34215 frac3860. 