đề 1
Ta có:
2P=2√yzx+2√yz+2√zxy+2√zx+2√xyz+2√xy
=1−xx+2√yz+1−yy+2√xz+1−zz+2√xy
=3−(√x2x+2√yz+√y2y+2√xz+√z2z+2√xy)
≤3−(√x+√y+√z)2x+y+z+2√xy+2√yz+2√zx (Cauchy-Schwarz)
=3−(√x+√y+√z)2(√x+√y+√z)2
=3−1=2
Suy ra P≤1
Vậy, maxP=1
Dấu bằng xảy ra khi x=y=z