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I came across this inequality in a graph theory book, couldn't figure how to prove it.

$$n\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}<ne^{-\ln(k+1)}.$$

$n$ and $k$ are both positive integers. (Amount of vertices and minimum degree if that matters.)

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    $\begingroup$ Are you familiar with the inequality $(1 + \tfrac1x)^{x} < e$? $\endgroup$ – Mees de Vries Oct 5 '18 at 9:57
  • $\begingroup$ No but if the proof involves it I would gladly learn it. $\endgroup$ – Fuseques Oct 5 '18 at 9:59
  • $\begingroup$ Correction: that inequality holds for positive $x$, and the reverse inequality for negative $x$. Now apply the inequality to $-n/(k+1)$... do you see how to go form there? $\endgroup$ – Mees de Vries Oct 5 '18 at 10:18
  • $\begingroup$ hmmmm no sorry... $\endgroup$ – Fuseques Oct 5 '18 at 10:43
  • $\begingroup$ Are you familiar with the inequality $1+x \leq e^x$? This is true for any $x \in \mathbb{R}$, and the equality holds if and only if $x = 0$. Geometrically, this follows from the fact that $e^x$ is strictly convex and $y=x+1$ is the tangent line at $x = 0$ $\endgroup$ – Sangchul Lee Oct 5 '18 at 11:03
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Starting from the well know inequality $\log(1+x)<x$ for $x\neq 0$ (proof, proof), we get \begin{align} \ln\left(1-\frac{k+1}{n}\right)&<-\frac{k+1}n\\ \frac n{k+1}\ln\left(1-\frac{k+1}{n}\right)&<-1\\ \frac{n\ln(k+1)}{k+1}\ln\left(1-\frac{k+1}{n}\right)&<-\ln(k+1)\\ \ln\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<-\ln(k+1)\\ \left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<e^{-\ln(k+1)}\\ n\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<ne^{-\ln(k+1)}\\ \end{align}

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  • $\begingroup$ Can you please explain how you derived the 4th line from the 3rd? is there some rule that if $$ab < c$$ then $$b^a < c$$? seems not right, so probably something else? $\endgroup$ – Fuseques Oct 6 '18 at 5:27
  • $\begingroup$ $a\ln (b)=\ln (b^a) $ $\endgroup$ – Fabio Lucchini Oct 6 '18 at 5:29
  • $\begingroup$ oh right, thank you! $\endgroup$ – Fuseques Oct 6 '18 at 5:49

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