$\prod_{i=1}^{n}x_i^{p_i}\le p_1x_1+\ldots +p_nx_n$ strict inequality when $p_i \neq p_j$ for $i \neq j$

Question

Let $x_i,p_i \in \mathbb{R}$ and $x_i,p_i>0$. I just showed that for $\sum_{i=1}^{n}p_i=1$ we have the following inequality:

$$\prod_{i=1}^{n}x_i^{p_i}\le p_1x_1+\ldots +p_nx_n$$

(I used Jensen's inequality for this)

Now the next question is why we do we have the strict inequality if all $x_i$ are different:

$$\prod_{i=1}^{n}x_i^{p_i}< p_1x_1+\ldots +p_nx_n$$ for $x_i\neq x_j$for $i \neq j$.

Can somebody answer this? Thanks in advance!

Answer 1

You mentioned that you used Jensen's inequality, which tells you that $$\prod_{i=1}^n x_i^{p_i} = \mathrm{exp} \left( \sum_{i=1}^n p_i \log x_i \right) \le \sum_{i=1}^n p_i \, \mathrm{exp}(\log x_i) = \sum_{i=1}^n p_ix_i$$

since $x \mapsto \mathrm{exp}(x)$ is convex.

But Jensen's inequality also tells you a necessary and sufficient condition for equality to hold—in this case, it says that equality holds if and only if $\log x_1 = \log x_2 = \cdots = \log x_n$. But this in turn holds if and only if $x_1 = x_2 = \cdots = x_n$.

So if even one $x_i$ is different from the other $x_j$, then you have strict inequality.