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Is there an expression or bound for $x,\varepsilon>0$ that satisfies $$x^{4(2x-1)}

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    Is $n$ given? Are you looking for a bound such that the inequality holds for **all** n? $n\in \mathbf N$? – 2017-02-05
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    @NiklasHebestreit of course at all large enough $n\in\Bbb N$. – 2017-02-05
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    Not clear to me what you're asking, as $x^{4x(2x-1)}\to \infty$ as $x\to \infty$. – 2017-02-05
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    @user254665 The OP is asking to set an upper bound on $x$ such that the inequality holds. – 2017-02-05
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    Taking logs, the expression is equivalent to $$\frac{4(2x-1)\log x}{(x-1)(2x-3)\epsilon} < \epsilon \log n.$$ Using very rough estimates: This fails if $x$ is sufficiently close to $0$, for example if $x < 1/4$ and $x < n^{-\epsilon}$. It also fails if $x$ is sufficiently close to $3/2$ with $x > 3/2$, for example if $$\frac{3}{2} < x < \frac{3}{2} + \frac{16 \log(3/2)}{2 \epsilon \log n}.$$ It holds for all $x$ large enough, for example it holds for all $x > 3$ when $1/e < \epsilon \log n$. I suppose the kind of answer you want depends on how sharp you need these estimates to be. – 2017-02-05

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