Summation inequality problem: $n +(1 + 1/2 + 1/3 +.....+1/n) > n(n + 1)^{1/n}$

Question

Let $S_{n} = 1 + 1/2 + 1/3 +.....+1/n,$ then prove that $$n + S_{n} > n(n + 1)^{1/n}.$$

My work: To show $1 + S_{n}/n > (n + 1)^{1/n}.$ I start with Cauchy-Schwartz inequality, $(1 + 2 + \cdots + n)(1 + 1/2 + 1/3 +.....+1/n)>n^2$ which implies $S_n/n > 2/(n + 1).$ But after adding $1$ to both sides I can not find the desired result. So is there any problem?

Answer 1

Hint: $$n+S_n = \frac{2}{1} + \frac{3}{2} + \frac{4}{3} + \cdots + \frac{n+1}{n}.$$ Now apply the AM-GM inequality.

Answer 2

$$1+\sum_{i=1}^n\frac{1}{i\cdot n}=\frac{n+\sum_{i=1}^n\frac{1}{i}}{n}$$ Now use the Inequality of arithmetic and geometric means and the fact that $\sum_{i=1}^n\frac{1}{i}>1$. It's only true for $n\geq 2$ otherwise you get $\geq$ instead of $>$