证明x三次方+y三次方+z三次方>=3xyz

你的题少条件
若x=y=-1,z=1
则你的不等式是-1>=3,不成立
所以应该是x>0,y>0,x>0

x³+y³+z³-3xyz
=(x³+3x^2y+3xy^2+y³+z³)-(3xyz+3x^2y+3xy^2)
=[(x+y)³+z³]-3xy(x+y+z)
=(x+y+z)(x^2+y^2+2xy-xz-yz+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+z^2+2xy-3xy-xz-yz)
=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

因为x>0,y>0,x>0
所以x+y+z>0

x²+y²+z²-xy-yz-xz
=(2x²+2y²+2z²-2xy-2yz-2xz)/2
=[(x²-2xy+y²)+(y²-2yz+z²)+(z²-2xz+x²)]/2
=[(x-y)²+(y-z)²+(z-x)²]/2>=0

所以相乘大于等于0‘
所以x³+y³+z³-3xyz>=0
x³+y³+z³>=3xyz

（x+y+z)^3=x^3+y^3+z^3+3x^2(y+z)+3y^2(x+z)+3z^2(x+y)+6xyz
已知x+y+z=0,得:x+y= -z ,x+z= -y, y+z= -x
0=x^3+y^3+z^3-3x^3-3y^3-3z^3+6xyz
2x^3+2y^3+2z^3=6xyz
即：x的三次方+y的三次方+z的三次方=3xyz

x^3+y^3+z^3-3xyz
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]/2
≥0
x^3+y^3+z^3≥3xyz