Prove that x^2 - 6x+13>=4 for real x numbers. Factor the expression A = x^2+4y^2+9-4xy+6x-12y. If x=2y, prove that A=9.

When solving a inequality, we have to determine the
interval of numbers for the expression to be positive or
negative.

In this case, we'll have to subtract 4 both
sides:

x^2 - 6x+13 -
4>=0

x^2 - 6x+ 9
>=0

We remark that we can write the expression above
as a binomial raised to square:

(x-3)^2 >=
0

The square of the expression x-3 is always positive for
any value of x, excepting x = 3 for the expression is
cancelling.

To factor the expresison A, we'll
re-write it:

A =
x^2+4y^2+9-4xy+6x-12y

A =x^2 - 4xy + 4y^2 + 6x - 12y +
9

We'll group the first 3
factors:

A = (x - 2y)^2 + 6(x - 2y) +
9

We'll put x - 2y = t

A = t^2
- 6t + 9

A = (t - 3)^2

We'll
put t = x - 2y

A = (x - 2y -
3)^2

We'll verify if x = 2y the value for A
is 9. For this reason, we'll substitute x by 2y:

A = (2y -
2y - 3)^2

We'll eliminate like
terms:

A = 9
q.e.d.