#### Question

If a, b are real numbers, find the minimum value of

a^2+ab+b^2-a-2b

Collected in the board: Quadratic function

Steven Zheng posted 14 hours ago

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

Steven Zheng posted 14 hours ago

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

Steven Zheng posted 14 hours ago

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