Inequality of $( y_1 \cdot y_2 \cdot \ldots \cdot y_M )\log ( y_1 \cdot y_2 \cdot \ldots \cdot y_M )$?

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I have the following form of a function,

$z=( y_1 \cdot y_2 \cdot \ldots \cdot y_M )\log( y_1 \cdot y_2 \cdot \ldots \cdot y_M )$

Is there any inequality relation between $z$ and $\prod\limits_{i = 1}^M y_i\log y_i$ for ${y_i} \in (0,1)$ or ${y_i} \in (1,\infty )$?

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Let $Y=\prod_{i=1}^{M} y_{i}$. Then $$ z=Y\log Y=Y\sum_{i=1}^{M}\log y_{i} $$

Let $w=\prod_{i=1}^{M} y_{i}\log y_{i}$. Then $w=Y\prod_{i=1}^{M}y_{i}$.

Since $Y>0$ comparing $z$ and $w$ is the same as comparing $a=\sum_{i=1}^{M}\log y_{i}$ and $b=\prod_{i=1}^{M}\log y_{i}$.

This will depend on much more than just $(0,1)$ and $(1,\infty)$.

For example if $y_{i}\in(0,1)$ and $M$ is even then clearly $a<0<b$.

But if $y_{i}\in (0,1)$ and $M$ is odd then we could have $a>b$ (for example take $y_i=\frac{1}{e^2}$ for all $i$ so $a=-2M$ and $b=-2^M$) or we could have $a<b$ (for example take $y_i=\frac{1}{e^{1/2}}$ for all $i$ so $a=-M/2$ and $b=-\frac{1}{2^{M}}$.

For $y_{i}\in (1,\infty)$ the odd/even dichotomy is not relevant, but we still have similar cases as in the odd case above.

So in general there are many cases and the critical points at $e$ and $\frac{1}{e}$ play a big role, in addition to $1$.

The main picture is that you are comparing the sum and product of $M$ real numbers: $\log y_{1}, \ldots, \log y_{M}$. So there is a lot of variation in the behavior.