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!
Suppose that $\epsilon \leq X \leq \frac{1}{4}$, how to show that $|X(1-6X)|\leq \frac{1}{4\epsilon^4}$?
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algebra-precalculus
inequality
3 Answers
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The inequality is rather trivial. $$|x(1-6x)|=|x||(1-6x)|\leq \frac{1}4\cdot \frac32=\frac38\leq64\leq \frac{1}{4\epsilon^4}$$
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0Small note: you typed cube, not fourth power. – 2017-01-09
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0Thank you for noticing the typo. I have fixed that – 2017-01-09
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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?
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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 $\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 $\quad 0<\epsilon\leq x\leq 1/4 \implies 0<\epsilon \leq 1/4\implies 1/4\epsilon^4\geq 256.$