Convex functions in integral inequality Let $\mu,\sigma>0$ and define the function $f$ as follows:
$$
f(x) = \frac{1}{\sigma\sqrt{2\pi}}\mathrm \exp\left(-\frac{(x-\mu)^2}{2\sigma ^2}\right)
$$
How can I show that
$$
\int\limits_{-\infty}^\infty x\log|x|f(x)\mathrm dx\geq \underbrace{\left(\int\limits_{-\infty}^\infty x f(x)\mathrm dx\right)}_\mu\cdot\left(\int\limits_{-\infty}^\infty \log|x| f(x)\mathrm dx\right)
$$
which is also equivalent to $\mathsf E[ X\log|X|]\geq \underbrace{\mathsf EX}_\mu\cdot\mathsf E\log|X|$ for a random variable $X\sim\mathscr N(\mu,\sigma^2).$

$\int x log \vert x \vert f(x) dx - \mu \int log \vert x \vert f(x) dx  = \int( x - \mu) log \vert x \vert f(x) dx  = \int x  log \vert x + \mu  \vert \phi(x) dx = \int_0^{\infty} x log(\vert \frac {\mu +x}{\mu - x}\vert )\phi(x) dx$     and the integrand is positive.  $\phi$ is symmetric is all that gets used.  
