<prompt> Acha x, y, na z ziwe nambari halisi chanya kama vile xy xz yz 1. Pata thamani ya chini zaidi ya 10x2 10y2 z2.  <response>: Tuseme usawa unatokea wakati x,y,z x_0,y_0,z_0. Ili kupata na kuthibitisha thamani ya chini, inaonekana tutalazimika kuweka pamoja baadhi ya ukosefu wa usawa kama vile ukosefu wa usawa.x2 y2 ge 2xy.Tukikumbuka kwamba usawa hutokea wakati x x_0 na y y_0, au fracxx_0 fracyy_0 1, tunaunda ukosefu wa usawafracx2x_02 fracy2y_02 ge frac2xyx_0 y_0.Kishafracy_02x_0 cdot x2 fracx_02y_0 cdot y2 ge xy.Vile vile,anzaalignfracz_02x_0 cdot x2 fracx_02z_0 cdot z2 ge xz, fracz_02y_0 cdot y2 fracy_02z_0 cdot z2 ge xz.endalignTukiongeza hizi, tunapatafracy_0 z_02x_0 cdot x2 fracx_0 z_02y_0 cdot y2 fracx_0 y_02z_0 cdot z2 ge xy xz yz.Tunataka kuongeza 10x2 10y2 z2, kwa hivyo tunataka x_0, y_0, na z_0 ili kuridhishafracy_0 z_0x_0 : fracx_0 z_0y_0 : fracx_0 y_0z_0 10:10:1.Hebuanzaaligny_0 z_0 10kx_0, x_0 z_0 10ky_0, x_0 y_0 kz_0.mwishoalignKishaanzaalignx_0 y_0 z_0 10k 1 x_0, x_0 y_0 z_0 10k 1 y_0, x_0 y_0 z_0 k 1 z_0.endalignAcha t x_0 y_0 z_0. Kisha x_0 fract10k 1, y_0 fract10k 1, na z_0 fractk 1, hivyofract10k 1 fract10k 1 fractk 1 t.Kwa hiyo,frac110k 1 frac110k 1 frac1k 1 1.Hii hurahisisha hadi 10k2 - k - 2 0, ambayo huchangia kama 2k - 15k 2 0. Kwa kuwa k ni chanya, k frac12.Kisha x_0 fract6, y_0 fract6, na z_0 frac2t3. Inabadilisha hadi xy xz yz 1, tunapatafract236 fract29 fract29 1.Kutatua, tunapata t 2, na thamani ya chini ya 10x2 10y2 z2 ni10 cdot fract236 10 cdot fract236 frac4t29 t2 4. 