# proof for $n(1-\frac{k+1}{n})^{n\ln(k+1)/(k+1)}<ne^{-\ln(k+1)}$

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)}

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

• Are you familiar with the inequality $(1 + \tfrac1x)^{x} < e$? – Mees de Vries Oct 5 '18 at 9:57
• No but if the proof involves it I would gladly learn it. – Fuseques Oct 5 '18 at 9:59
• 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 5 '18 at 10:18
• hmmmm no sorry... – Fuseques Oct 5 '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 5 '18 at 11:03

Starting from the well know inequality $$\log(1+x) for $$x\neq 0$$ (proof, proof), we get \begin{align} \ln\left(1-\frac{k+1}{n}\right)&<-\frac{k+1}n\\ \frac n{k+1}\ln\left(1-\frac{k+1}{n}\right)&<-1\\ \frac{n\ln(k+1)}{k+1}\ln\left(1-\frac{k+1}{n}\right)&<-\ln(k+1)\\ \ln\left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&<-\ln(k+1)\\ \left(1-\frac{k+1}{n}\right)^{n\ln(k+1)/(k+1)}&
• Can you please explain how you derived the 4th line from the 3rd? is there some rule that if $$ab < c$$ then $$b^a < c$$? seems not right, so probably something else? – Fuseques Oct 6 '18 at 5:27
• $a\ln (b)=\ln (b^a)$ – Fabio Lucchini Oct 6 '18 at 5:29