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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.)

amWhy
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Carpet4
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    Are you familiar with the inequality $(1 + \tfrac1x)^{x} < e$? – Mees de Vries Oct 05 '18 at 09:57
  • No but if the proof involves it I would gladly learn it. – Carpet4 Oct 05 '18 at 09:59
  • 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? – Mees de Vries Oct 05 '18 at 10:18
  • hmmmm no sorry... – Carpet4 Oct 05 '18 at 10:43
  • 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 $x = 0$ – Sangchul Lee Oct 05 '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}

Fabio Lucchini
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