I am stuck with the following problem. Given the Gaussian mixture distribution $f(\cdot)$
$$ f(x) = \frac{a}{\sqrt{2\pi\sigma_1^2}}e^{-\frac{x^2}{2\sigma_1^2}}+\frac{1-a}{\sqrt{2\pi\sigma_2^2}}e^{-\frac{(x-\mu)^2}{2\sigma_2^2}} $$ where $a\in (0,0.5)$, $\mu>0$ and $\sigma_1 \neq \sigma_2$ so that $f(x)$ is skewed. The mixtures of $f(x)$, $$ f_1(x) = p\cdot f(x-c) + (1-p)\cdot f(x+c), $$ $$ f_2(x) = \frac{1}{2}\cdot f(x-c) +\frac{1}{2}\cdot f(x+c) $$ where $c$ is a constant. Prove that there exist a $p\in(0,1)$ so that the following inequality holds:
$$ -\int_{-\infty}^{\infty}f_1(x)\log f_1(x)dx>-\int_{-\infty}^{\infty}f_2(x)\log f_2(x)dx $$ where the LHS is the entropy of $f_1(x)$ and the RHS is the entropy of $f_2(x)$.Any suggestions? Thanks for your time in advance.
Define $$h(p) := -\int_{-\infty}^{\infty}f_1(x,p)\log f_1(x,p)\,dx $$ and show that for some $p\in(0,1)$, $$h(p) > h(1/2). $$
This can be done by showing that $h$ has a maximum at some $p \in(0,1)\backslash\{1/2\}$.
Hints:
Find $p$ such that $\frac{d}{dp}h(p) = 0$ and $\frac{d^2}{dp^2}h(p) < 0$.
$\int_{-\infty}^{\infty}\left(\frac{\partial}{\partial\,p}f_1(x,p)\right)dx = 0.$
$\frac{d}{dp}h(p) = -\int_{-\infty}^{\infty}\left(\frac{\partial}{\partial p}f_1(x,p)\right)\log f_1(x,p)\,dx $
$\frac{\partial^2}{\partial p^2}f_1(x,p) = 0.$
$\frac{d^2}{dp^2}h(p) = -\int_{-\infty}^{\infty}\left(\frac{\frac{\partial}{\partial p}f_1(x,p)}{f_1(x,p)} \right)^2f_1(x,p)\, dx.$