A tricky inequality: $n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.$ Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$ be the $n-$th harmonic number,
it is not difficult to prove that:
(1) $n(n+1)^{\frac{1}{n}}<n+H_n,$ for $n>1$; (use AM-GM inequality)
(2) $(n-2)n^{\frac{1}{n-2}}>n-H_n$, for $n>2$. (Hint: $n=3$ is obvious,
when $n>3$, $n-(n-2)n^{\frac{1}{n-2}}<2<H_n$)
So from (1), we know
$$H_n>n(n+1)^{\frac{1}{n}}-n, n>1;\ (3)$$
from (2), we know
$$H_n>n-(n-2)n^{\frac{1}{n-2}},n>2.\ (4)$$
It seems inequality $(3)$ is better than $(4)$, that is to say,
$$\boxed{n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.}$$
My concern is whether there are tricky proofs for above inequality without complicated derivative calculations.
Any helps, hints and comments will welcome.
 A: By Bernoulli
$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}=\frac{n}{\left(1+\frac{1}{n+1}-1\right)^{\frac{1}{n}}}+\frac{n-2}{\left(1+\frac{1}{n}-1\right)^{\frac{1}{n-2}}}\geq$$
$$\geq\frac{n}{1-\frac{1}{n+1}}+\frac{n-2}{1+\frac{1}{n-2}\cdot\frac{1-n}{n}}=n+1+\frac{n(n-2)^2}{n^2-3n+1}=$$
$$=2n+\frac{1}{n^2-3n+1}>2n$$
A: Hint
You can use a majorization theorem as follows :
Let $a\geq b>0 $ and $c\geq d>0$ if we have :
$a\geq c $ $\, \operatorname{and}$ $ab\geq cd $ then we have :
$$a+b\geq c+d$$
Here you can take $c=d=n$ and $a=n(n+1)^{\frac{1}{n}}$,$\,$$b=(n-2)n^{\frac{1}{n-2}}$
Then it's easier to use derivative .The theorem I used is a direct consequence of the Karamata's inequality .
A: Using that for $n>1$
$$1+n\geq3>e>\left(1+\frac1n\right)^n\implies (n+1)^{\frac{1}{n}}>1+\frac1n.$$
We have
$$ n(n+1)^{\frac{1}{n}}>n+1,\ n>1.$$
And then for $n>3$,
$$(n-2)n^{\frac{1}{n-2}}>(n-2)(n-1)^{\frac{1}{n-2}}>(n-2)+1=n-1,$$
when $n=3$, we also have
$$(n-2)n^{\frac{1}{n-2}}=3>2=n-1$$
therefore
$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>n+1+n-1=2n,n\geq 3.$$
