Let $X$ be a random variable. I am trying to prove the bound $$ \log \mathbb{E} |X|-\mathbb{E}\log |X|\leq\mathbb{E}\left[\frac{1}{2}\left(1+|X|\right)(\log|X|)^2\right]$$
What I tried so far:
We know that $x\leq1+\log(x)+\frac{1}{2}(\log(x))^2(x+1)$ so that $$ \log \mathbb{E} |X|-\mathbb{E}\log |X| \leq \log \mathbb{E} \left[1+\log|X|+\frac{1}{2}\left(1+|X|\right)(\log|X|)^2\right]-\mathbb{E}\log |X|$$
Since $\mathbb{E}1=1$ $\mathbb{E}$ is linear and $\log$ is sub-additive (by being concave), we find $$ \log \mathbb{E} |X|-\mathbb{E}\log |X| \leq \log \mathbb{E} \left[\frac{1}{2}\left(1+|X|\right)(\log|X|)^2\right]+\log\mathbb{E}\left[\log|X|\right]-\mathbb{E}\log |X|$$
Now use the fact that $\log(x)\leq x$ to obtain the result.
Question
The book where I read this says that one also needs the bound $1+x\leq e^x$ but I don't see where. So either the above is false, or there's a typo in the book? What's the right answer?