# Suppose that $\epsilon \leq X \leq \frac{1}{4}$, how to show that $|X(1-6X)|\leq \frac{1}{4\epsilon^4}$?

Suppose that for $\epsilon>0$, $\epsilon \leq X \leq \frac{1}{4}$, I'd like to show that $|X(1-6X)|\leq \frac{1}{4\epsilon^4}$. However, I just can't get it. I've tried the product inequality and also triangle inequality, but am not getting the bound right. Can someone offer some guidance? thanks!

The inequality is rather trivial. $$|x(1-6x)|=|x||(1-6x)|\leq \frac{1}4\cdot \frac32=\frac38\leq64\leq \frac{1}{4\epsilon^4}$$

• Small note: you typed cube, not fourth power. Commented Jan 9, 2017 at 12:24
• Thank you for noticing the typo. I have fixed that Commented Jan 9, 2017 at 15:29

Note that for $0<\epsilon\leq 1/4$ then $1/(4\epsilon^4)\geq 4^3=64$.

On the other hand, if $f(x)=|x(1-6x)|$ then for $x\in [0,1/4]$, $$f(x)\leq\max(f(1/12),f(1/4))=\max(1/24,1/8)=1/8.$$

So, your inequality seems to be quite trivial. Are you sure that you wrote the statement correctly?

Complete the square. $F(x)=x(1-6x)=-6((x-1/12)^2-1/144)=-6(x-1/12)^2+1/24.$ Now

$0<\epsilon \leq x\leq \frac {1}{4}\implies 0<x\leq \frac {1}{4}\implies -\frac {1}{12}<x-\frac {1}{12}\leq \frac {1}{4}-\frac {1}{12}=\frac {1}{6}\implies$

$\implies |x-\frac {1}{12}|\leq \frac {1}{6}\implies 0\leq 6(x-\frac {1}{12})^2\leq \frac {1}{6}\implies$

$-\frac {1}{8}\leq F(x)\leq \frac {1}{24} \implies |F(x)|\leq \frac {1}{8}.$

And $\quad 0<\epsilon\leq x\leq 1/4 \implies 0<\epsilon \leq 1/4\implies 1/4\epsilon^4\geq 256.$