令x-1=y+12=z−23=t，则x=t+1，y=2t-1，z=3t+2，于是x2+y2+z2=（t+1）2+（2t-1）2+（3t+2）2=t2+2t+1+4t2+1-4t+9t2+4+12t=14t2+10t+6，∵x，y，z均为非负数，∴x-1≥-1，y+12≥12，z−23≥-23，∵x-1=y+12=t，∴y≥1...