A tricky inequality: $n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>2n,\ n\geq3.$

Question

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.

Answer 1

By Bernoulli $$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}=\frac{n}{\left(1+\frac{1}{n+1}-1\right)^{\frac{1}{n}}}+\frac{n-2}{\left(1+\frac{1}{n}-1\right)^{\frac{1}{n-2}}}\geq$$ $$\geq\frac{n}{1-\frac{1}{n+1}}+\frac{n-2}{1+\frac{1}{n-2}\cdot\frac{1-n}{n}}=n+1+\frac{n(n-2)^2}{n^2-3n+1}=$$ $$=2n+\frac{1}{n^2-3n+1}>2n$$

Answer 2

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 .

Answer 3

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

$$n(n+1)^{\frac{1}{n}}+(n-2)n^{\frac{1}{n-2}}>n+1+n-1=2n,n\geq 3.$$