2(a²+b²+1)-2(ab+a+b)
=2a²+2b²+2-2ab-2a-2b
=(a²-2ab+b²)+(a²-2a+1)+(b²-2b+1)
=(a-b)²+(a-1)²+(b-1)²
≥0
所以2(a²+b²+1)≥2(ab+a+b)
a²+b²+1≥ab+a+b(当a=b=1时取等号)
2(a²+b²+1)-2(ab+a+b)
=2a²+2b²+2-2ab-2a-2b
=(a²-2ab+b²)+(a²-2a+1)+(b²-2b+1)
=(a-b)²+(a-1)²+(b-1)²
≥0
所以2(a²+b²+1)≥2(ab+a+b)
a²+b²+1≥ab+a+b(当a=b=1时取等号)