Answer
Solution set: $(-\infty,\infty)$
Work Step by Step
Follow the "Procedure for Solving Polynomial lnequalities",\ p.412:
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$4x^{2}+1 \geq 4x$
$4x^{2}-4x+1 \geq 0$
$f(x)=4x^{2}-4x+1 \quad $...factor the trinomial...
find factors of $4(1)=4$ that add to $-4:$
($-2$ and $-2$ )
$4x^{2}-4x+1=4x^{2}-2x-2x+1= \quad$ ... in pairs ...
$=2x(2x-1)-(2x-1)=(2x-1)(2x-1)$
(we also could/should have recognized this square of a difference)
$f(x)=(2x-1)^{2} \geq 0 $
...
There is no need to follow the steps any further, because the above inequality "asks":
"When is a square of a real number greater or equal to zero?"
The answer: "Always."
That is, for all x $\in \mathbb{R}$
Solution set: $(-\infty,\infty)$
