Question
If a, b are real numbers, find the minimum value of
a^2+ab+b^2-a-2b
Answer 1
Let
A = a^2+ab+b^2-a-2b
(1)
Rearrange the equation to the one in terms of a
a^2+(b-1)a+b^2-2b-A = 0
Then, the discriminant
\Delta = (b-1)^2 - 4(b^2-2b-A) \geq 0
3b^2-6b-4A-1\leq 0
(2)
Since f(x) = 3x^2-6x-4A-1 opens upwards, the vertex coordinates must be below origin if satisfying (1).
Then using the vertex formula for y coordinate
\Big( \dfrac{-b}{2a},\dfrac{4ac-b^2}{4a} \Big)
\dfrac{4\cdot 3(-4A-1)-(-6)^2 }{4\cdot 3 } \leq 0
Solving the inequality yields
A\geq -1
(3)
Substituting to (2) results in the equation
3b^2-6b-4(-1)-1= 0
b^2-2b+1 = 0
b =1
(4)
Substitute (3) and (4) to (1), we get
a = 0
(5)
Therefore
The minimum value of a^2+ab+b^2-a-2b is 1 when a=0 and b=1
Answer 2
Let
A = a^2+ab+b^2-a-2b
(1)
Let
a = u-v
(2)
b = u+v
(3)
A = a^2+ab+b^2-a-2b
=(u-v)^2+(u+v)^2+(u-v)(u+v)-(u-v)-2(u+v)
=2u^2+2v^2+u^2-v^2 -u+v-2u-2v
=3u^2-3u+v^2-v
=3(u-\dfrac{1}{2} )^2-\dfrac{3}{4}+(v-\dfrac{1}{2} )^2-\dfrac{1}{4}
=3(u-\dfrac{1}{2} )^2+(v-\dfrac{1}{2} )^2-1
Therefore
When u = \dfrac{1}{2} and v = \dfrac{1}{2} , that is, a = 0, b=1
a^2+ab+b^2-a-2b has minimum value of -1