Is the proof approach for $\sqrt[n]{\Pi_{j=1}^n x_j} \leq \frac{1}{n}\sum_{j=1}^n x_j$ futile? I need to prove for $n \in \mathbb{N} \text{\\} \{ 0\}$ and $x_j \in \mathbb{R^{+}}$ that the inequality for geometric and arithmetic mean holds.
I use induction.
First, Let n=1, then it is clear that $x_1^{1/1} \leq x_1$
Therefore, step 2, assume the inequality holds for some $k=n$.
Step 3, I need to show that inequality holds $k+1$ from the fact that it holds for $k$. Therefore I write
$$\sqrt[k]{\Pi_{j=1}^k x_j} \leq \frac{1}{k}\sum_{j=1}^k x_j$$
It is clear that $\frac{1}{k}\sum_{j=1}^kx_j < \frac{1}{k}\sum_{j=1}^{k+1}x_j$, thus
$$\sqrt[k]{\Pi_{j=1}^k x_j} \leq \frac{1}{k}\sum_{j=1}^{k+1} x_j$$
It is also clear that $(x_1)^{(1/(k+1))}(x_2)^{(1/(k+1))}...(x_k)^{(1/(k+1))} < (x_1)^{(1/k)}(x_2)^{(1/k)}...(x_k)^{(1/k)}$, thus
$$\sqrt[k+1]{\Pi_{j=1}^k x_j} \leq \frac{1}{k}\sum_{j=1}^{k+1} x_j$$
I do not know exactly how to proceed now, I had the following idea, which incredibly dirty and probably won't lead anywhere. Define the difference between the RHS of the inequality above with the one where we divide each term by $1/k+1$ as the following: 
$$\Delta^* = \frac{\sum_{j=1}^{k+1}x_j}{k(k+1)}$$
The difference between the LHS of the same inequality and the one where we multiply by additional term $x_{k+1}$ is more dirty:
$$\Delta_*=\Pi_{j=1}^{k+1}x_j^{1/(k+1)}-\Pi_{j=1}^{k}x_j^{1/(k+1)}=\Pi_{j=1}^kx_j^{1/(k+1)}(x_{k+1}^{1/(k+1)}-1)$$
And so in theory if I prove that:
$$\Pi_{j=1}^kx_j^{1/(k+1)}(x_{k+1}^{1/(k+1)}-1) \leq \frac{\sum_{j=1}^{k+1}x_j}{k(k+1)}$$
then I have proceed it by induction. But it feels as if I have made matters worse. I would be interested to see if I can proceed with this approach any furhter?

Hint:
Let
$A_n
=\frac1{n}\sum_{k=1}^n x_k
$
and
$G_n
=\sqrt[n]{\prod_{k=1}^n x_k}
$.
Write
$A_{n+1}$
and
$G_{n+1}$
in terms of
$x_{n+1},
A_{n}$,
and
$G_{n}$.
Use the induction hypothesis that
$G_n
\le A_n$
and see what that gets you.
It might make it easier
if you assume that
$x_{n+1}$
is extreme
(i.e., either
$x_{n+1} \le x_k$
or
$x_{n+1} \ge x_k$
for
$1 \le k \le n$).
