Question

设a,b为实数，那么a^2+ab+b^2-a-2b的最小值

Answer 1

令a=u+v,b=u-v则
a²+ab+b²-a-2b=(u+v)²+(u+v)(u-v)+(u-v)²-(u+v)-2(u-v)
=3u²+v²-3u+v
=3(u²-u)+(v²+v)
=3(u-1/2)²+(v+1/2)²-3/4-1/4
=3(u-1/2)²+(v+1/2)²-1
≥-1