设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx

已知x+y+z=2,xy+yz+zx=-5,求x²+y²+z²的值 已知x+y+z=1,x²+y²+z²=2求xy+yz+zx (1/x+1/y+1/z)×(xy)/(xy+yz+zx) 已知2x-y-5z=0,x-2y+2z=0,求x²+y²+z²/xy+yz+zx的值. 设x,y,z≥0,x+y+z=3,证明:√x+√y+√z≥xy+yz+zx 证明 (x+y+z)^2>3(xy+yz+zx) 分解因式：xyz-yz-zx-xy+x+y+z-1 xy+yz+zx=1,x,y,z>=0 xy+yz+zx=1,求x√yz+y√zx+z√xy 1.已知4(xy-zx-y²+yz)=-z²+2zx-x²,求z-zy+x-3的值 已知X,Y,Z都是整数且xy+yz+zx=1,求证x+y+z>=根号3 正实数x，y，z满足9xyz+xy+yz+zx=4，求证：