# estimating $\prod_{i=1}^k a_i\leq \max_i a_i^{p_i}$, where $\sum_i 1/p_i=1$

Let $$1 such that $$\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 ...

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 $$p_i$$...

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 $$b_i=p_i \ln \,a_i$$.

• Concise and quick! – GNUSupporter 8964民主女神 地下教會 Apr 30 at 9:56
• What if $a_i<1$? Then $b_i$ is negative and the argument doesn't work, or does it? – Pink Panther Apr 30 at 10:08
• @PinkPanther Thank you for your comment. Taking non-negative $b_i$'s was totally unnecessary in my argument. – Kavi Rama Murthy Apr 30 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

https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means#Weighted_AM%E2%80%93GM_inequality

We have:

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