Show that x^{2}- 12x + 11 + 18ln(x) > 0, for x > 1.

My attempt:

I start by evaluating x^{2}- 12x + 11 = p.

The expression equals zero at x = 1, x = 11. It is positive for all x > 11 and negative on 11 > x > 1.

Since ln(x) > 0 for every x > 1, and the expression p is positive when x > 11; the inequality holds for x > 11.

I'm guessing the next step is to show that (x-11)(x-1) > -18ln(x) for 11 > x > 1.

But I don't know how to proceed from here