Multivariable Calculation- Entropy of A Function I'm having trouble understanding the following calculation we've learned in class.
The objective was to prove a log-Sobolev inequality.
We did it this way:
We know that $$ \int_\Omega \sqrt{ | \nabla(\epsilon f^2 ) | ^2 + (D \cdot I(\epsilon f^2) )^2 } d\mu \geq D I\left(\int_\Omega \epsilon f^2 d \mu\right) $$ for some constant $D>0$ ($  \Omega$ is a region, $\mu$ is some measure on it, and $I(\epsilon ) = \sqrt{2} \epsilon \sqrt{\log(1/\epsilon)} $  when $\epsilon \to 0 $ ). 
The lecturer then said that by taking $ \epsilon \to 0 $ , we get: 
$$ \frac{D^2 } {2} \left( \int_\Omega f^2 \log f^2 d\mu -\int_\Omega f^2 d\mu \cdot \log \int_\Omega f^2 d\mu \right)\leq \int _\Omega | \nabla f| ^2 d\mu $$  (The LHS is excatly the entropy of $f^2$)
Can someone help me understand the calculation?
Thanks !
 A: The admitted inequality gives 
$$\int_{\Omega}\sqrt{\varepsilon^2|\nabla (f^2)|^2+2D^2\varepsilon^2f^4\left(-\log(\varepsilon f^2)\right)}d\mu\geq D\sqrt 2\varepsilon\int_{\Omega}f^2d\mu\sqrt{-\log\left(\int_{\Omega}\varepsilon f^2\right)},$$
hence 
$$\int_{\Omega}\sqrt{4f^2|\nabla f^2|-2D^2f^4\log(\varepsilon f^2)}d\mu\geq D\sqrt 2\int_{\Omega}f^2d\mu\sqrt{-\log\left(\int_{\Omega}\varepsilon f^2\right)}.$$
The LHS is 
$$\leq \sqrt{\int_{\Omega}f^2d\mu}\sqrt{\int_{\Omega}(4|\nabla f|^2-2D^2f^2\log(\varepsilon f^2))d\mu}.$$
We get 
$$\sqrt{\int_{\Omega}(4|\nabla f|^2-2D^2f^2\log(\varepsilon f^2))d\mu}\geq D\sqrt 2\sqrt{\int_{\Omega}f^2d\mu}\sqrt{-\log\left(\int_{\Omega}\varepsilon f^2\right)},$$
and taking the squares,
$$\int_{\Omega}(4|\nabla f|^2-2f^2D^2\log(\varepsilon f^2))d\mu\geq 2D^2\int_{\Omega}f^2d\mu\left(-\log\left(\int_{\Omega}\varepsilon f^2\right)d\mu\right).$$
This gives 
$$\int_{\Omega}|\nabla f|^2d\mu\geq \frac{D^2}2\left(\int_{\Omega}f^2\log(\varepsilon f^2)d\mu-\int_{\Omega}f^2d\mu\log\int_{\Omega}(\varepsilon f^2)d\mu\right).$$
When $\varepsilon \to\color{red}1$, it gives the wanted inequality.  
