设a,b为实数,则a²+ab+b²-a-2b的最小值是

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设a,b为实数,则a²+ab+b²-a-2b的最小值是

a²+ab+b²-a-2b
=a^2+a(b-1)+b^2-2b
=[a+(b-1)/2]^2+(3/4)b^2-(3/2)b-1/4
=[a+(b-1)/2]^2+(3/4)(b-1)^2-1,
当a+(b-1)/2=0,b=1,即a=0,b=1时它取最小值-1.

a^2+ab+b^2-a-2b=1/4(a+2b)^2 +3/4a^2 -(a+2b)
=[1/2(a+2b) -1 ]^2 + 3/4a^2 -1≥0+0-1=-1

a²+ab+b²-a-2b
=a²+ab-a+b²-2b
=a²+a(b-1)+(b²-2b+1)-1
=a²+a(b-1)+(b-1)²-1
=[a²+a(b-1)+(b-1)²/4]+3(b-1)²/4-1
=[a-(b-1)/2]²+3(b-1)²/4-1
≥-1