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 $p_i$...


3 Answers 3


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 $b_i=p_i \ln \,a_i$.

  • $\begingroup$ Concise and quick! $\endgroup$ Commented Apr 30, 2019 at 9:56
  • $\begingroup$ What if $a_i<1$? Then $b_i$ is negative and the argument doesn't work, or does it? $\endgroup$ Commented Apr 30, 2019 at 10:08
  • 2
    $\begingroup$ @PinkPanther Thank you for your comment. Taking non-negative $b_i$'s was totally unnecessary in my argument. $\endgroup$ Commented Apr 30, 2019 at 10:14

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 $\log x$: $$\sum_{i=1}^k\frac{1}{p_i}\log x_i \leq \log \left(\sum_{i=1}^k\frac{1}{p_i}x_i \right)\leq \log \left(\max_i x_i \cdot \sum_{i=1}^k\frac{1}{p_i} \right) = \log\max_i x_i $$


You can use weighted AM-GM inequality here


We have:

$$\prod\limits_i a_i \leq \sum\limits_i \frac{a_i^{p_i}}{p_i}\leq \max{a_i^{p_i}}\sum\limits_i\frac{1}{p_i}=\max{a_i^{p_i}}$$


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