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I'm looking at a proof for obtaining an inequality and in the final line there is this inequality.

$$\frac{C}{2^{m-1}-1}(2+\log(2^{m-1}-1))\le \frac{Cm}{2^m}$$ where $C$ is some constant and $m\in \mathbb{N}$. I can't figure out how to show this inequality. I would greatly appreciate any help.

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    What constraints are there on $C$? Is it positive? – 2017-02-11
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    @BrevanEllefsen yes positive – 2017-02-11
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    Then I think your inequality is backwards – 2017-02-11
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    I don't think the inequality holds when $C$ is positive. In that case, for large positive $m$, $\frac{C}{2^{m-1}-1}(2 + \log(2^{m-1}-1)) \approx \frac{C}{2^{m-1}}(1+m) \ge \frac{Cm}{2^m}$. – 2017-02-11
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    @NilabroSaha exactly. I think the OP has this backwards. Of course, even then some weird stuff happens for small $m$. The inequality is more of a thing just for large $m$, as you note – 2017-02-11
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    @BrevanEllefsen I uploaded the proof where I got this inequality from and I need this bound for the bottom proof in obtaining a contradiction. Can you perhaps tell me where this proof goes wrong? – 2017-02-11
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    The inequality clearly doesn't apply for $m=1$. Graphing the LHS and RHS seems to suggest that the inequality in its present form, with $C$ positive, doesn't hold for any $m\in \Bbb{N} -{1}$. – 2017-02-11
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    Of course, the reverse inequality does hold, judging from the graphs. – 2017-02-11
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    @NilabroSaha I can confirm that your analysis of the graphs is correct from doing some algebra on the inequality. We can write it as $$\frac{4+2\ln \left(x-1\right)}{1+\log _2\left(x\right)} \le \frac{x-1}{2x}$$ which is clearly false by taking limits. – 2017-02-11
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    In fact, for all $x>3$ we have that the LHS is monotonically decreasing toward $2\ln(2)$ and the RHS is moronically increasing toward $1/2$. – 2017-02-11
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    Yes that's what I noticed. Moreover, the graphs seemed to me like the Maxwellian distribution. – 2017-02-11
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    I know this is severely deviating from the question, but is there a reason why the graphs look like the Maxwellian distribution. I actually don't know the equation for the distribution, so maybe the LHS and the RHS resemble it? – 2017-02-11

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