TITLE: If $x$ is real and $p=\frac{3(x^2+1)}{2x-1}$, prove that $p^2-3(p+3)\ge0$
QUESTION [7 upvotes]: If $x$ is real and $p=\frac{3(x^2+1)}{2x-1}$, prove that $p^2-3(p+3)\ge0$
So far, I have tried to substitute the value of $p$ into $p^2-3(p+3)$ and tried to see if I could find anything that would lead me to the conclusion that the expression is greater than or equal to zero.
What I have ended up with is $\frac{9\{(x^2+1)^2-(2x-1)(x^2+1)-(2x-1)^2\}}{(2x-1)^2}$
I do not see how this could lead me to proving what the questions asks however.

REPLY [10 votes]: Note that
$$(2x-1)p=3(x^2+1)\Rightarrow 3x^2-2px+3+p=0.$$
Now consider the discriminant.