Can somebody gave me an idea about how to prove the following inequality: $$(n-1)^{n-1}(3n+1) \geq n^{n-2}(1+n^2), n \geq 2, n \in \mathbb{N}?$$
I tried to use logarithms to get the expression: $$ (n-1) \lg (n-1) +\lg (3n+1) \geq (n-2) \lg n + \lg(1+n^2),$$ but for example from here I don't know how to do it $$n \lg \frac{n-1}{n} \geq \lg \left( \frac{1+n^2}{3n+1} \cdot \frac{n-1}{n^2}\right).$$
Maybe I started wrong. Any idea? Thanks!
Dividing by $n^n$ this is equivalent, when $n>1,$ to $$\frac{1}{\left(1+\frac1{n-1}\right)^{n-1}}\left(3+\frac1n\right)\geq 1+\frac1{n^2}$$
Or:
$$3+\frac1n\geq \left(1+\frac1{n^2}\right) \left(1+\frac1{n-1}\right)^{n-1}.\tag1$$
Since $\left(1+\frac1{n-1}\right)^{n-1}<e,$ and if $n>3,$ $$1+\frac1{n^2}<\frac3e,$$ so when $n>3,$ the right side of $(1)$ is less than $3.$
So you only need to check $n=1,2,3.$
You get the stronger inequality:
$$3(n-1)^{n-1}>n^{n-3}(n^2+1)$$
This assumes we know, for $m>0$ that:
$$\left(1+\frac1m\right)^m<e\tag2.$$
There are a number of ways of proving this, the most elementary is to use binomial theorem to expand the left side of $(2)$ and compare the terms to $$e=\sum_{k=0}^{\infty}\frac1{k!}.$$
I suppose the logarithm won’t get you anywhere. Divide both sides by $n^n$ and you get $(1-1/n)^{n-1} (3+1/n) >= 1 + 1/n^2$.
Now the left side is close to 3/e, and the right side close to 1. Actually the left side has a limit of 3/e = 3/2.71828… and the limit of the right side is 1.
On second thought… change the right hand side to $ \log ((1 -1/n + 1/n^2 - 1/n^3 ) / ( 3+1/n)) ≈ - \log 3$ and use the Taylor series for log 1-x on the left side, that should get you there.
We have that
$$(n-1)^{n-1}(3n+1) \geq n^{n-2}(1+n^2) \iff \left(1-\frac1n\right)^{n-1}\frac{3n^2+n}{n^2+1} \geq 1$$
which is true, indeed for $n\ge 2$
$$\frac{3n^2+n}{n^2+1} =3+\frac{n-3}{n^2+1} \ge 2.8\ge e$$
$$\left(1-\frac1n\right)^{n-1} \ge \frac1e$$
Refer also to the related
