√(x - 3) ≥ 1/(x - 3) help on how to solve please!!!

Before solving the inequality, we'll impose constraints of
existence of square root.

x - 3
>=0

We'll add 3 both sides:

x
>= 3

The interval of admissible values of x is [3 ,
+infinite)

We'll raise to square both sides, to get rid of the
square root:

(x - 3) ≥ 1/(x -
3)^2

We'll subtract 1/(x - 3)^2 both
sides:

(x - 3) - 1/(x - 3)^2 ≥  0

We'll
multiply by (x - 3)^2 both sides:

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

We'll apply the formula of difference of
cubes:

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

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

We'll square raise and we'll combine like
terms:

(x-3)^3 - 1 = (x-4)(x^2 - 5x +
7)

(x-4)(x^2 - 5x + 7) >=0

A
product is positive if and only if the 2 factors are both negative or both
positive.

Case
1)

x-4>0

x>4

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 value of x.

From both
inequalities, the interval for admissible values for x is (4; +
infinite).

Case
2)

x-4<0

x<4

x^2
- 5x + 7 < 0

But the expression is always
positive, for any value of x, so x belongs to empty set.