Inequality in differential entropy In the book on "Network Information Theory" by El Gamal, there is a question to choose the correct relation ($\geq,\leq,=$) for the following:
Let $X$ be a continuous random variable. Let $Y\sim N(0,1)$ (Standard normal distributed) and $Y$ is independent of $X$. Let $a \geq 1$. Then what is the relation between $h(X+aY)$ and $h(X+Y)$?
Here $h(.)$ is the differential entropy operator defined as follows: If $X$ has pdf $f_X$, then
$$h(X) = -\int_{-\infty}^\infty f_X(x)\log_2(f_X(x))dx$$
My guess is that it should be $\geq$. For instance, if $X$ is normal with variance $P$, then I have showed that it is $\geq$. For general $X$, i thought of manipulating the entropy terms with some clever conditioning but I didn't get anywhere. As a last resort, I even considered expanding out each term and trying to do a comparison but again, to no avail. 
I am sure there is a clever trick/transformation here that would give the answer, but I am unable to see it. Hence my request for help.
Edit: You are welcome to use the Entropy Power Inequality if it helps. In fact any property of differential entropy may be used here. Also as $Y$ is standard normal, $h(Y) = \frac{\log_2(2\pi e )}{2}$.
 A: Hint: Let $Z_a=X+aY$, and $w(a)=h(Z_a)=h(X+aY)$, in nats. Then show that $w(a)$ increases, using de Bruijn’s identity :
$$ \frac{\partial }{\partial t} h(X+\sqrt{t}\,Y)=\frac{1}{2}J(X+\sqrt{t}\,Y)\ge 0$$
where $J(\cdot)$ is the Fisher information (see eg Cover & Thomas, Theorem 17.7.2).
A: Found a nice alternative proof that uses the property of standard normal. Essentially, given two zero mean Gaussian random variables, they are equal in distribution iff their variances are equal.
Assume $a>1$ as $a=1$ case is trivial. Thus given $Y \in \mathcal{N}(0,1)$
$$aY \stackrel{d}{=} Y_1 + \sqrt{a^2-1}Y_2$$
where $Y_1,Y_2$ are iid $\mathcal{N}(0,1)$, independent of $X$. Also we observe that if $Y\stackrel{d}{=}Z$, and $Z$ is independent of $X$, then
$$h(Y)=h(Z), \quad h(X+Y) = h(X+Z)$$
This is because the densities $f_{X+Y}$ and $f_{X+Z}$ are equal. From this we get
\begin{align}
h(X+aY) &= h(X+Y_1+\sqrt{a^2-1}Y_2) \\
&\geq h(X+Y_1+\sqrt{a^2-1}Y_2|Y_2) \\
&= h(X+Y_1|Y_2)\\
&= h(X+Y_1)\\
&= h(X+Y)
\end{align}
This gives the desired result without using anything too complicated.
