Verified answer

By squaring both sides, we obtain:

x - 3 ≥ 1/(x - 3)^2.

Then, assuming that x ≠ 3, we can multiply both sides by (x - 3)^2 to get:

(x - 3)^3 ≥ 1.

By cube rooting both sides:

[(x - 3)^3]^(1/3) ≥ 1^(1/3)

==> x - 3 ≥ 1

==> x ≥ 4.

I hope this helps!

√x -3 ≥ 1/(x - 3) you must solve

√x -3 - 1/(x - 3) ≥ 0 because if multiply by (x-3) alien solution appear !

[(x-3)^(3/2) - 1] /(x-2) ≥ 0

(√(x-3) - 1)((√(x-3))^2 +√(x-3) +1 ) /(x-2) ≥ 0

but ((√(x-3))^2 +√(x-3) +1 ) > 0 for any real x.

Then you must study only sign of

(√(x-3) - 1) /(x-2) ≥ 0

because x-3 must be positive (under √ sign) then x≥3

for x ≥ 3 the nominator is positive then you must study sign of

√(x-3) - 1

if x = 3 no sense because of 1/(x-3)

for x in (3,4) negative value

for x = 4 null value

for x in (4,+infinity) positive value