Theo giả thiết, 1x+1y+1z=1x+y+z.

Suy ra yz+xz+xyxyz=1x+y+z.

yz+xz+xyx+y+z=xyz

yzx+y+z+xzx+y+z+xyx+y+z=xyz

xyz+y2z+yz2+x2z+xyz+xz2+x2y+xy2+xyz=xyz

x2z+2xyz+y2z+yz2+xz2+x2y+xy2=0

zx+y2+z2x+y+xyx+y=0

x+yzx+y+z2+xy=0

x+yxz+yz+z2+xy=0

x+yxz+xy+yz+z2=0

x+yxy+z+zy+z=0

x+yy+zx+z=0

Suy ra x+y=0  hoặc y+z=0  hoặc x+z=0.

⦁ Nếu x+y=0  thì x=−y,  khi đó x2023=−y2023.

Ta có 1x2023+1y2023+1z2023=1−y2023+1y2023+1z2023=1z2023;

1x2023+y2023+z2023=1−y2023+y2023+z2023=1z2023.

Do đó 1x2023+1y2023+1z2023=1x2023+y2023+z2023.

⦁ Nếu y+z=0  hoặc x+z=0,  chứng minh tương tự ta cũng có

1x2023+1y2023+1z2023=1x2023+y2023+z2023.