Let $f: [0, \infty)^n \rightarrow \mathbb{R}$
$$f(x)=\frac{(x_1x_2...x_n)^\frac{1}{n+1}}{1+x_1+...+x_n}$$
Show that the function achieves its global maximum at $(1,1,...,1) \in \mathbb{R^n}$ and deduce that
For all $y_1, ... , y_{n+1} \in [0, \infty)$
$$(y_1...y_{n+1})^\frac{1}{n+1}\leq \frac{1}{n+1} \sum_{i=1}^{n+1}y_i$$
I have shown that the function's global maximum is at $(1,...,1)$, after that I get that $f(x)\leq \frac{1}{n+1}$, so
$$(x_1x_2...x_n)^\frac{1}{n+1}\leq \frac{1}{n+1}(1+x_1+...+x_n)$$
I fail to see how I can show the proposition for a $n+1$. It seems like I should vary the $1$, but I fail to see how I can achieve the AM GM inequality that way. Any hints are welcome.
Let $y_1, \ldots, y_{n+1} \in [0, +\infty)$ and assume $y_{n+1} \ne 0$.
Then
\begin{align} (y_1\ldots y_{n+1})^{\frac1{n+1}} &= y_{n+1}\left(\frac{y_1}{y_{n+1}}\cdots\frac{y_n}{y_{n+1}}\right)^{\frac1{n+1}}\\ &\le y_{n+1} \frac{1 + \frac{y_1}{y_{n+1}} + \cdots + \frac{y_n}{y_{n+1}}}{n+1}\\ &= \frac{y_1 + \cdots + y_n + y_{n+1}}{n+1} \end{align}