# Proof using the AM-GM inequality.

I've been studying for my upcoming exams and I came across this exercise in the AM-GM inequality section:

If $a_n = \sqrt[n]{a}$ , $a \in \mathbb{R}$ , $a > 0$ , $n \in \mathbb{N}$ then prove that:

$$\dfrac{na}{1 + na} < a_n < 1 + \dfrac{a}{n}.$$

I've been trying to prove this for the past few hours but I am completely stuck at this point. Any help would be much appreciated!

A start: Consider the numbers $1,1,1,\dots,1$ ($n-1$ of them) and $a$. Their arithmetic mean is $\frac{n-1+a}{n}$ and their geometric mean is $\sqrt[n]{a}$. Now note that $1+\frac{a}{n}\gt \frac{n-1+a}{n}$.