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. 
A: Consider the function
$$f(x)=x(1-ax) \implies f'(x)=1-2ax\implies f''(x)=-2a$$ The first derivative cancels at $x_*=\frac 1 {2a}$ and $$f\left(\frac{1}{2 a}\right)=\frac{1}{4 a}$$ The point $x_*$ is a maximum if $a>0$ by the second derivative test and a minimum otherwise.
A: We can also proceed as follow
$$x(1 - ax) \le \frac{1}{4a} \iff ax^2- x+\frac{1}{4a}\ge 0$$
which holds when $a>0$ since
$$\Delta=1-4\cdot \frac{1}{4a}=1-1=0$$
A: Without any derivative, just high-school theory of quadratic equations:
A quadratic polynomial (with real coefficients) has a global extremum at the arithmetic mean of its roots. Further, this extremum is a maximum if its leading coefficient is negative, a minimum if the leading coefficient is positive.
So here, the extremum is attained  at $\dfrac1{2a}$ and the leading coefficient is $-a$. This extremum is a maximum if $a>0$, a minimum if $a<0$ and it is equal to
$$\frac1{2a}\biggl(1-a\,\frac1{2a}\biggr)=\frac1{4a}.$$
A: If you already know that $x(1-x) \le 1/4$ for all $x$, then it also holds for $ax$. Divide the inequation you get by $a$ and you obtain the result you wanted.
