I want to show the following: given a measure space $(X,\mu)$ and $f,g$ $\mu$-measurable functions on $X$, $$\int_X |f(x)g(x)| d\mu(x) \leq \frac{1}{2}\int_{|f(x)| \leq 1} |f(x)|^2 d\mu(x) + \frac{1}{2} \int_{|g(x)|\leq 1} |g(x)|^2 d\mu(x) + \int_{|f(x)| > 1} \left(|f(x)|\log |f(x)| + \frac{1}{2} \right) d\mu(x) + \int_{|g(x)| > 1} \left(e^{|g(x)| - 1} - \frac{1}{2}\right) d\mu(x)$$
To that end, I have shown the following:
Let $\phi(s)$ be continuous and strictly increasing for $s \geq 0$ with $\phi(0) = 0$. Then $$\int_X |f(x)g(x)| d\mu(x) \leq \int_X \int_0^{|f(x)|} \phi(s) ds d\mu(x) + \int_X \int_0^{|g(x)|} \phi^{-1}(s) ds d\mu(x)$$ where $\phi^{-1}$ is the function inverse of $\phi$.
So it seems to me that proving the above inequality is simply a matter of choosing the right function $\phi$, which appears to be some kind of logarithm. I can't seem to find it, however. In particular, my estimates always fail to include the 1/2's on the first two integrals, and I'm not sure how to choose a function that will be nice enough. Any input or ideas would be greatly appreciated!
The function you want is $$\varphi(s) = \begin{cases} s,\quad & 0\le s\le 1 \\ 1+\log s, \quad &s>1\end{cases}$$ Reason: since we want to write an integral of the form $\int_X \Psi(|f(x)|)\,d\mu(x)$ as $\int_X \int_0^{|f(x)|} \phi(s) \,ds \,d\mu(x) $, the function $\phi$ should be the derivative of $\Psi$.
Things to check: $$\int_0^t\varphi(s)\,ds = \begin{cases} t^2/2,\quad & 0\le t\le 1 \\ 1/2+t\log t, \quad &t>1\end{cases}\tag{1}$$ $$\varphi^{-1}(s) = \begin{cases} s,\quad & 0\le s\le 1 \\ e^{s-1} , \quad &s>1\end{cases} \tag{2}$$ $$\int_0^t\varphi^{-1}(s)\,ds = \begin{cases} t^2/2,\quad & 0\le t\le 1 \\ 1/2+e^{t-1} , \quad &t>1\end{cases}\tag{3}$$ All are straightforward.