Tham khảo bài này nha:
Cho x;y;z không âm,thoảvx+y+z=3.Chứng minh rằng :2(x2+y2+z2)+xyz≥72(x2+y2+z2)+xyz≥7
Ta có:
(3−2x)(3−2y)(3−2z)≤(9−2x−2y−2z)327=1(3−2x)(3−2y)(3−2z)≤(9−2x−2y−2z)327=1
⇔8xyz≥−28+12(xy+yz+zx)⇔8xyz≥−28+12(xy+yz+zx)
⇔xyz≥−144+32(xy+yz+zx)⇔xyz≥−144+32(xy+yz+zx)
Ta có:
2(x2+y2+z2)+xyz≥54(x2+y2+z2)+34(x+y+z)2−144≥512(x+y+z)2+34(x+y+z)2−144