<prompt> Acha x, y, na z ziwe nambari halisi zisizo hasi kama vile x2 2y2 5z2 22. Pata thamani ya juu zaidi yaxy xz yz.  <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.Kwa kuwa tumepewa hiyo x2 2y2 5z2 22, tunataka x_0, y_0, na z_0 ili kukidhifracy_0 z_0x_0 : fracx_0 z_0y_0 : fracx_0 y_0z_0 1:2:5.Hebuanzaaligny_0 z_0 kx_0, x_0 z_0 2ky_0, x_0 y_0 5kz_0.mwishoalignKishaanzaalignx_0 y_0 z_0 k 1 x_0, x_0 y_0 z_0 2k 1 y_0, x_0 y_0 z_0 5k 1 z_0.endalignAcha t x_0 y_0 z_0. Kisha x_0 fractk 1, y_0 fract2k 1, na z_0 fract5k 1, hivyofractk 1 fract2k 1 fract5k 1 t.Kwa hiyo,frac1k 1 frac12k 1 frac15k 1 1.Hii hurahisisha hadi 10k3 - 8k - 2 0, ambayo ni 2k - 15k2 5k 1 0. Kwa kuwa k lazima iwe chanya, k 1.Kisha x_0 fract2, y_0 fract3, na z_0 fract6. Inabadilisha hadi x2 2y 2 5z2 22, tunapatafract24 frac2t29 frac5t236 22.Kutatua, tunapata t 6, na thamani ya juu zaidi ya xy xz yz nifract2 cdot fract3 fract2 cdot fract6 fract3 cdot fract6 frac1136 t2 11. 