Q:

Is $x(1 - 2x) \le \frac{1}{8}$ and further, is $x(1 - ax) \le \frac{1}{4a}$

It is clear that $x(1-x) \le \frac{1}{4}$
Does it likewise follow that $x(1-2x) \le \frac{1}{8}$? 
Here's my reasoning:
(1)  For $x < \frac{1}{4}$, $x(1-2x) < \frac{1}{8}$
(2)  For $\frac{1}{4} < x < \frac{1}{2}$, $x(1-2x) < \frac{1}{8}$
(3)  For $\frac{1}{2} < x$, $x(1-2x) < 0$
Further, can this be generalized to $x(1-ax) \le \frac{1}{4a}$
Since:
(1) For $x < \frac{1}{2a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(2) For $\frac{1}{2a} < x < \frac{1}{a}$, $x(1-ax) < \left(\frac{1}{2a}\right)\left(\frac{1}{2}\right) = \frac{1}{4a}$
(3) For $\frac{1}{a} < x$, $x(1-ax) < 0$
Are both of these observations correct?   Is only one correct?   Is there an exception that I am missing? 

A:

The first it's
$$16x^2-8x+1\geq0$$ or
$$(4x-1)^2\geq0.$$
The second for $a>0$ it's
$$4a^2x^2-4ax+1\geq0$$ or $$(2ax-1)^2\geq0.$$
For $a<0$ the second inequality is reversed. 