Weighted AM-GM inequality with reciprocal of natural numbers and exponents

Question

If $n$ is a positive integer, show that $$\left(1-\frac{1}{n}\right)^n < \left(1-\frac{1}{n+1}\right)^{n+1}$$

So well if we take common denominators and simplify we get,

\begin{align} &\frac{(n-1)^n}{n^n} < \frac{(n)^{n+1}}{(n+1)^{n+1}} \\ \Rightarrow &(n^2-1)^n \cdot (n+1) < (n^2)^n \cdot n \\ \end{align}

And since $(n^2)^n$ is much greater $(n^2-1)^n$ i can see that the transformed version of the inequality is true. However i couldn't come up with an actual method to prove it. I also couldn't figure out appropriate weights for the original inequality. We are supposed to use Weighted\Regular AM-GM-HM. Also maybe if we define $f(x)=\left(1-\dfrac{1}{x}\right)^x$ and show that is increasing we will be done ? However is that doable/viable approach? The main method however is supposed to be WAM-WGM-WHM. Thanks.

Answer 1

Let $a=1-1/n$ and $b=1-1/(n+1)$. Then $0\le a<b$. Then $$b^{n+1}-a^{n+1}=(b-a)(a^n+a^{n-1}b+\cdots+b^n)>(n+1)(b-a)a^n =\frac{a^n}n.$$ So $$b^{n+1}>a^n\left(a+\frac1n\right)=a^n.$$

Answer 2

For $n=1$ it's obvious.

But for $n>1$ by your work we need to prove that: $$\left(\frac{n^2}{n^2-1}\right)^n>1+\frac{1}{n}$$ or $$\left(1+\frac{1}{n^2-1}\right)^{n^2}>\left(1+\frac{1}{n}\right)^n,$$ which is true because $$\left(1+\frac{1}{n^2-1}\right)^{n^2}>e>\left(1+\frac{1}{n}\right)^n.$$