<prompt> Tafuta nambari halisi kubwa zaidi c kama vile x_12 x_22 dots x_1012 geq cM2wakati wowote x_1,x_2,ldots,x_101 ni nambari halisi ambazo x_1x_2cdotsx_1010 na M ni wastani wa x_1,x_2,ldets,x_101.  <response>: Kwa kuwa ukosefu wa usawa daima ni kweli kwa M 0, inatosha kuzingatia kesi M neq 0.Kwa c mahususi na kwa nakala yoyote x_1, dots, x_101 inayokidhi masharti, nakala -x_1, dots, -x_101 inakidhi masharti pia, kwa hivyo tunaweza kudhani kuwa M 0. Hatimaye, tunaweza kudhani kuwa x_1 le x_2 le dots le x_101, ili M x_51.Tunataka kupata c kubwa zaidi kwamba ukosefu wa usawa x_12 x_22 dots x_1012 ge cx_512 huwa daima, ambapo x_1 le x_2 le nukta le x_101 na x_1 x_2 vitone x_101 0. Kwa hivyo, tukirekebisha thamani ya x_51, tunapaswa kuandika ukosefu wa usawa unaopunguza x_1 2 x_22 vitone x_1012.Ili kulinganisha masharti ya upande wa kushoto na x_512, tunashughulikia masharti x_12 x_22 dots x_502 na x_51 2x_522dotsx_1012 tofauti.Na Cauchy-Schwarz, 1 1 dots 1x_12x_22dotsx_502 ge x_1x_2dotsx_50 2,kwa hivyo x_12 x_22 dots x_502 ge tfrac150kushotox_1x_2dotsx_50kulia 2.Tuna x_1x_2dotsx_50 -x_51-x_52 -dots - x_101le -51x_51 kwa sababu x_ 51 le x_52 le le dots le x_101. Tangu x_51 0, zote x_1 x_2 dots x_50 na -51x_51 ni hasi, kwa hivyo tunaweza kuandika beginaligned x_12x_22dotsx_502 ge tfrac150 x_1x_2dots x_502 getfrac150 kushoto-51x_51kulia2 tfrac51250 x_ 512. endalignedKwa upande mwingine, kwa kuwa 0 x_51 le x_52 le dots le x_101, tunayo x_512 x_ 522 dots x_1012 ge 51x_512.Kuweka haya yote pamoja kunatoa x_12 x_22 dots x_50 2 x_512 x_522 vitone x_1012 ge kushototfrac51250 51kulia x_ 512 tfrac515150 x_512.Usawa unashikilia wakati x_1 x_2 dots x_50 -tfrac5150 na x_51 x_52 dots x_101 1, kwa hivyo jibu ni tfrac515150. 