estimating $\prod_{i=1}^k a_i\leq \max_i a_i^{p_i}$, where $\sum_i 1/p_i=1$ Let $1<p_1,\dots,p_k<\infty$ such that $\sum_{i=1}^k\frac{1}{p_i}=1$.
Moreover, let $a_1,\dots,a_k\geq 0$.
I have to show that 
$$\prod_{i=1}^k a_i\leq\max_i a_i^{p_i}$$
I want to use induction over $k$, but I am struggling with it.
For $k=2$ I only got to here:
Since $p_1$ and $p_2$ are Hölder conjugates, we have
$\frac{1}{p_1}+\frac{1}{p_2}=1\Leftrightarrow p_1=\frac{p_2}{p_2-1}$.
I have been experimenting with the case when $a_1 a_2>a_2^{p_2}$, but I didn't get anywhere
    \begin{align*}
  1<a_2^{p_2-1}a_1^{-1}
  \Leftrightarrow
  1<a_2 a_1^{-1/(p_2-1)}
  \Leftrightarrow
  1<a_2\frac{a_1^{p_1}}{a_1}
\end{align*}
...
I thought maybe case distinction between the four cases $a_i\leq 1$ and $a_i>1$ could work, but I don't see how this works, since we also have to take care of the $p_i$...
 A: We may suppose $a_i \neq 0$ for any $i$. Let $q_i=\frac 1 {p_i}$. Then $\sum q_i  b_i \leq \max \{b_i\}$ for any set of real numbers $b_i$ because $\sum q_i=1$. Hence $\prod_i e^{b_iq_i} \leq e^{\max {b_i}}=\max e^{b_i}$. Put $b_i=p_i \ln \,a_i$. 
A: Assuming $a_i > 0$ for $i=1,\ldots , k$ (otherwise the inequality is trivial) you may set 


*

*$a_i = x_i^{\frac{1}{p_i}}$
So, it is enough to show for $x_i > 0$
$$ \prod_{i=1}^k x_i^{\frac{1}{p_i}}\leq\max_i x_i$$
But this follows immediately by the concavity and monotonicity of $\log x$:
$$\sum_{i=1}^k\frac{1}{p_i}\log x_i \leq \log \left(\sum_{i=1}^k\frac{1}{p_i}x_i \right)\leq \log \left(\max_i x_i \cdot \sum_{i=1}^k\frac{1}{p_i} \right) = \log\max_i x_i $$
A: You can use weighted AM-GM inequality here
https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Weighted_AM%E2%80%93GM_inequality
We have:
$$\prod\limits_i a_i \leq \sum\limits_i \frac{a_i^{p_i}}{p_i}\leq \max{a_i^{p_i}}\sum\limits_i\frac{1}{p_i}=\max{a_i^{p_i}}$$
