Show that $1+\frac{1}{a}\bigg(\sum_{n=1}^a\frac{1}{n}\bigg)\gt \sqrt[a]{(a+1)}$

Question

How can I show that for all $a\in \mathbb{N}\ge 2$ that:

$$1+\frac{1}{a}\bigg(\sum_{n=1}^a\frac{1}{n}\bigg)\gt \sqrt[a]{(a+1)}$$

My first thought was to find a closed formula for the $n$th partial sum of the harmonic series, which there really isn't one. Not sure what else to do.

Answer 1

$1 +\frac 1a\left( \sum_\limits{n=1}^a \frac 1n\right)\\ \frac 1a\left(a + \sum_\limits{n=1}^a \frac 1n\right)\\ \frac 1a\left(\sum_\limits{n=1}^a 1+ \frac 1n\right)\\ \frac 1a\left(\sum_\limits{n=1}^a \frac {n+1}n\right)$

AM-GM inequality says

$\frac 1a\left(\sum_\limits{n=1}^a \frac {n+1}n\right) \ge \left(\prod_\limits{n=1}^a \frac {n+1}n\right)^\frac 1a$

And, we only have equality if all the terms in the average are identical.

$\prod_\limits{n=1}^a \frac {n+1}n = (2)(\frac 32)(\frac 43)\cdots(\frac {a}{a-1})(\frac {a+1}{a}) = a+1$

$\frac 1a\left(\sum_\limits{n=1}^a \frac {n+1}n\right) > (a+1)^\frac 1a$