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

Question

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?

Answer 1

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.

Answer 2

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

$$\Delta=1-4\cdot \frac{1}{4a}=1-1=0$$

Answer 3

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.

Answer 4

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 $$\frac1{2a}\biggl(1-a\,\frac1{2a}\biggr)=\frac1{4a}.$$

Answer 5

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.