<prompt> Acha x, y, na z ziwe nambari halisi chanya. Pata thamani ya juu zaidi yafracxyz1 5x4x 3y5y 6zz 18.  <response>: Kwanza, tunafanya maneno katika denominator kufanana. Kwa mfano, tunaweza kuzidisha kipengele 4x 3y kwa frac54 na pia tunazidisha nambari kwa frac54, ambayo hutupatiafracfrac54 xyz1 5x5x frac154 y5y 6zz 18.Sisi basi zidisha kipengele 5y 6z kwa frac34 na nambari, ambayo inatupafracfrac1516 xyz1 5x5x frac154 yfrac154 y frac9 2 zz 18.Mwisho, tunazidisha kipengele z 18 na frac92 na nambari, ambayo hutupatia.fracfrac13532 xyz1 5x5x frac154 yfrac154 y frac9 2 zfrac92 z 81.Hebu a 5x, b frac154 y, na c frac 92 z. Kisha x frac15 a, y frac415 b, na z frac29 c , hivyo usemi unakuwafracfrac120 abc1 aa bb cc 81.Na AM-GM,anzaalign1 a 1 fraca3 fraca3 fraca3 ge 4 sqrt4fraca3 27, a b a fracb3 fracb3 fracb3 ge 4 sqrt4fraca b3 27, b c b fracc3 fracc3 fracc3 ge 4 sqrt4fracb c3 27, c 81 c 27 27 27 ge 4 sqrt4c cdot 273.mwishoalignKisha1 aa bb cc 81 ge 4 sqrt4fraca327 cdot 4 sqrt4 fraca b327 cdot 4 sqrt4fracb c327 cdot 4 sqrt4c cdot 273 256abc ,hivyofracfrac120 abc1 aa bb cc 81 le fracfrac120 abc256 abc le frac15120.Usawa hutokea wakati a 3, b 9, na c 27, au x frac3 5, y frac125, na z 6, kwa hivyo thamani ya juu zaidi ni frac15120. 