How do you solve this inequation: √x -3 >= 1/(x - 3)?

First, we'll impose the constraints of existence of the square
root:

x - 3>
=0

x>=3

Now, we'll solve the
inequality by raising to square both sides:

(x - 3) >=1/(x -
3)^2

Now, we'll subtract 1/(x - 3)^2 both
sides:

(x - 3) - 1/(x - 3)^2
>=0

We'll multiply by (x-3)^2 the
inequality:

(x - 3)^3 -
1>=0

We'll solve the difference of cubes using the
formula:

a^3 - b^3 = (a-b)(a^2 + ab +
b^2)

(x - 3)^3 - 1 = (x - 3  -1)[(x-3)^2 + x - 3 +
1]

We'll combine like terms inside
brackets:

(x - 4)[(x-3)^2 + x - 3 + 1]
>=0

A product is zero if both factors have the same sign.
We'll get 2 cases to analyze:

Case
1)

x-4>=0

x>=4

x^2
- 6x + 9 + x - 2 >=0

x^2 - 5x + 7
>=0

x1 = [5+sqrt(25 -
28)]/2

Since delta = 25-28 = -3<0, the
expression

x^2 - 5x + 7 > 0 for any
x.

The common solution is the interval [4,
+infinity).

Case
2)

x-4=<0

x=<4

x^2
- 5x + 7 <0 impossible, because x^2 - 5x + 7 >0 for any value of
x.

The solution of the inequality is the interval [4,
+infinity).