Here's the question you clicked on:
frx
Show that \[x ^{2}+18\ln(x)+11>12x \] for x>1, I have no idea how to do the 18ln(x) messes things up for me...
damn i got cut off!!!
here is my guess start with \[x ^{2}-12x+18\ln(x)+11\]
take the derivative, show that it is positive on \(x>1\)
that will mean your function is increasing, and since it is greater than zero at 1 (i am assuming it is, i didn't check) that means it will always be greater than zero
actually at 1 it is zero but that doesn't change the solution