Young's generalizated inequality If $a_{1},a_{2},\dots,a_{n}$ are non-negative numbers, $p_{1},p_{2},\dots,p_{n}, \ p_{i}>1,$ for all $i=1,\dots,n$ and  $\sum_{i=1}^{n}\dfrac{1}{p_{i}}=1,$ then $$\prod_{i=1}^{n} a_{i} \leq \sum_{i=1}^{n} \frac{a_{i}^{p_{i}}}{p_{i}}.$$
Could anyone help me? For case $n=2$ is classic and I know. But I couldn't generalize.
 A: $\log$ is concave, therefore using Jensen's inequality we have
$$ \log\left(\prod_{i=1}^n a_i\right)=\sum_{i=1}^n\log(a_i)=\sum_{i=1}^n\frac{\log(a_i^{p_i})}{p_i}\leqslant \log\left(\sum_{i=1}^{n}\frac{a_i^{p_i}}{p_i}\right)$$
Thus $\displaystyle\prod_{i=1}^n a_i\leqslant\sum_{i=1}^n\frac{a_i^{p_i}}{p_i}$
A: A nice and not so difficult way to prove Young's generalized inequality is by induction on $n \geq 2$.
The base case for $n = 2$ is classic and you said you already know how to prove it so let's assume the desired result is valid for $n=2$.
Now, suppose that the inequality is true for some $n \geq 2$. That is, given any nonnegative real numbers $b_{1}, \ldots , b_{n}$ and any real numbers $q_{1}, \ldots, q_{n}$ such that $q_{i} > 1$ for $i = 1, \ldots , n$ and $\sum_{i=1}^{n} \frac{1}{q_{i}} = 1$, then $\prod_{i=1}^{n} b_{i} \leq \sum_{i=1}^{n} \frac{b_{i}^{q_{i}}}{q_{i}}$.
For the inductive step, assume that $a_{1}, \ldots , a_{n}, a_{n+1}$ are nonnegative real numbers and $p_{1}, \ldots , p_{n}, p_{n+1}$ are real numbers satisfying the conditions $p_{i} > 1$ for $i = 1, \ldots , n+1$ and $\sum_{i=1}^{n+1} \frac{1}{p_{i}} = 1$. We want to show that $\prod_{i=1}^{n+1} a_{i} \leq \sum_{i=1}^{n+1} \frac{a_{i}^{p_{i}}}{p_{i}}$. Let $a= \prod_{i=1}^{n} a_{i} \geq 0$ and $x = \sum_{i=1}^{n} \frac{1}{p_{i}} = 1 - \frac{1}{p_{i+1}}$. Notice that $x \in (0,1)$. Then $y = \frac{1}{x} > 1$, $p_{n+1} > 1$, and $\frac{1}{y} + \frac{1}{p_{n+1}} = x + \frac{1}{p_{n+1}} = 1$. Since $a$ and $a_{n+1}$ are nonnegative real numbers, by Young's inequality (which is the base case for $n=2$), we know that:
$\prod_{i=1}^{n+1} a_{i} = a \cdot a_{n+1} \leq \frac{a^{y}}{y} + \frac{a_{n+1}^{p_{n+1}}}{p_{n+1}}$.
Therefore, it is enough to prove that $\frac{a^{y}}{y} \leq \sum_{i=1}^{n} \frac{a_{i}^{p_{i}}}{p_{i}}$. For this purpose observe that $a_{1}^{y}, \ldots , a_{n}^{y}$ are nonnegative real numbers, $\frac{p_{1}}{y}, \ldots , \frac{p_{n}}{y}$ are real numbers greater than $1$ (because clearly $0 < \frac{1}{p_{i}} < x$ for $i = 1, \ldots , n$) and $\sum_{i=1}^{n} \left( \frac{p_{i}}{y} \right)^{-1} = y \sum_{i=1}^{n} \frac{1}{p_{i}} = yx = 1$. Our induction hypothesis lets us conclude that:
$a^{y} = \prod_{i=1}^{n} a_{i}^{y} \leq \sum_{i=1}^{n} \frac{(a_{i}^{y})^{\frac{p_{i}}{y}}}{\frac{p_{i}}{y}} = y \sum_{i=1}^{n}\frac{a_{i}^{p_{i}}}{p_{i}}$,
which leads us immediately to the desired inequality.
