What are all real values of a if f(x)=18x^2-lnx >= a, x>0 ?

The first thing to do is to analyze the monotony of the
function f(x). For this reason, we'll differentiate the
function:

f'(x) = 36x -
1/x

We'll put f'(x) = 0.

36x -
1/x = 0

(36x^2 - 1)/x = 0

We
notice that we have a difference of 2 squares at
numerator:

(36x^2 - 1) = (6x - 1)(6x +
1)

f'(x) = 0 if and only if  (6x - 1)(6x + 1) =
0.

6x - 1 = 0 => x =
1/6

6x + 1 = 0

x =
-1/6

The derivative is negative, over the range (0;1/6) and
it is positive over the range (1/6 ; +infinite).

That means
that the function is decreasing over the interval (0;1/6] and it is increasing over the
range [1/6 ; +infinite).

So, the point f(1/6) is a local
minimum point for the function.

Therefore,
f(x)>=f(1/6)>=a

f(1/6) = 18/36 - ln(1/6) =
1/2 + ln 6

So, a =< 1/2 + ln
6

All real values of "a" are located in the
interval (-infinite ; 1/2 + ln6].