<s>[INST] 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.
 [/INST] 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
$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>n+1+n-1=2n,n\geq 3.$$
</s>