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If $a \ge 0$, then how can I find the constant $c >0$ such that for $x \ge 1$ and $0 \le b \le x/2$, $$ \frac{1}{(x-b)^a} \le c \frac{1}{x^a}\;? $$

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Manipulate the inequality so that $x$ appears only on one side. Then maximize this function of $x$, –  Hans Engler Dec 1 '12 at 16:58
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You want to find some $c>0$ such that $$ c\frac{(x-b)^a}{x^a}\geq1\ \ (1). $$ The main observation here is that $$ c\frac{(x-b)^a}{x^a}\geq c\frac{(x-x/2)^a}{x^a} = \frac{c}{2^a} $$ Hence, if we require that $c\geq2^a$, then (1) is surely satisfied.

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