换元.可设a=x+y,b=x-y.(x,y∈R).则M=a²+ab+b²-a-b+(1/2)=(x+y)²+(x+y)(x-y)+(x-y)²-(x+y)-(x-y)+(1/2)=3x²-2x+y²+(1/2)=3[x-(1/3)]²+y²+(1/6).====>M=3[x-(1/3)]²+y²+(1/6)≥1/6.等号仅当x=1/3,y=0时取得,故Mmin=1/6.此时,a=b=1/3.【注：换元的目的即是消除ab项.】