Vì x,y>0 có P≥((x2+y2)22+z4)(2x2y2+1z4)≥((x+y)48+z4)(32(x+y)4+1z4)P≥((x2+y2)22+z4)(2x2y2+1z4)≥((x+y)48+z4)(32(x+y)4+1z4)
Đặt t=(x+y)4t4.(t∈[0;1])
Suy ra p≥(t8+1)(32t+1)=5+t8+32t=5+(t8+18t)+(32t−18t)=5+18(t+1t)+(32−18).1t≥5+28+(32−18)=2978p≥(t8+1)(32t+1)=5+t8+32t=5+(t8+18t)+(32t−18t)=5+18(t+1t)+(32−18).1t≥5+28+(32−18)=2978
Dấu bằng xảy ra x=y=z2