= Solution
For an $n$-dimensional <random vector> with a density and finite <differential entropy> in bits, define its <entropy power> by $N(X)=(2\pi e)^{-1}2^{2h(X)/n}$. The <entropy power inequality> states that for independent such <random vectors>,
$$
\boxed{N(X+Y)\geq N(X)+N(Y),\quad\text{equivalently}\quad
2^{2h(X+Y)/n}\geq2^{2h(X)/n}+2^{2h(Y)/n}.}
$$
Here the sum's <differential entropy> must be defined, with $+\infty$ permitted.
Since $g'(x)>0$, the <mean value theorem> makes $g$ strictly increasing. Its image is an interval $I$, and its inverse on $I$ has <derivative> $1/g'(g^{-1}(y))$. The density <change of variables> gives
$$
p_{g(X)}(y)=
\begin{cases}
\dfrac{p_X(g^{-1}(y))}{g'(g^{-1}(y))},&y\in I,\\
0,&y\notin I.
\end{cases}
$$
This does not require $g$ to map onto all of $\mathbb R$. Taking logarithms at $y=g(X)$ and then <expected values> proves the <differential entropy under an increasing transformation>:
$$
\boxed{h(g(X))=-\mathbb E\log_2p_X(X)+\mathbb E\log_2g'(X)
=h(X)+\mathbb E\log_2g'(X).}
$$
The two finite expectations in the assumptions justify splitting the expectation and imply that the transformed <differential entropy> is finite.
To derive the product inequality, assume that $h(Y_i)$ and $\mu_i=\mathbb E\log_2Y_i$ are finite and that the <differential entropy> of the logarithmic sum is defined. Set $Z_i=\ln Y_i$, using natural logarithms for this transformation while measuring <differential entropy> in bits. On the positive half-line the <derivative> of $\ln y$ is $1/y$, so the same density argument gives
$$
h(Z_i)=h(Y_i)-\mu_i.
$$
The $Z_i$ are <independent random variables>, and the one-dimensional <entropy power inequality> yields
$$
2^{2h(Z_1+Z_2)}\geq2^{2h(Z_1)}+2^{2h(Z_2)}.
$$
Since $Y_1Y_2=\exp(Z_1+Z_2)$, another <change of variables> gives
$$
h(Y_1Y_2)=h(Z_1+Z_2)+\mathbb E\log_2(Y_1Y_2)
=h(Z_1+Z_2)+\mu_1+\mu_2.
$$
Multiplying the <entropy power inequality> by $2^{2(\mu_1+\mu_2)}$ proves the <multiplicative entropy power inequality>:
$$
\boxed{2^{2h(Y_1Y_2)}
\geq 2^{2\mu_2}2^{2h(Y_1)}+2^{2\mu_1}2^{2h(Y_2)},\qquad
\alpha_1=2^{2\mu_2},\quad\alpha_2=2^{2\mu_1}.}
$$
The printed positivity and density assumptions alone do not ensure that the quantities in this last formula exist. For a concrete example, let $T$ have the standard <Cauchy distribution> and put $Y=e^T$. Then
$$
p_Y(y)=\frac{1}{\pi y(1+(\ln y)^2)},\qquad y>0,
$$
is a density, but $\mathbb E\log_2Y=(\log_2e)\mathbb ET$ is undefined because its positive and negative parts both diverge. Two independent copies satisfy the printed hypotheses but leave $\alpha_1,\alpha_2$ undefined. The product inequality therefore needs the finiteness or well-definedness conditions used above.
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