Let $f,g,h$ be measurable functions on a common measure space, $(X, \mathcal{A}, \mu)$ with $\mu(X) = 1$. If $|f|,|g|,|h| \leq 1$, show that,
$$\left|\int_X fh d\mu - \int_Xghd\mu\right| \leq 1 - \int_Xfgd\mu$$
My first instinct was to use Cauchy-Schwarz, but to no avail,
\begin{eqnarray}\left|\int_X fh d\mu - \int_Xghd\mu\right| &\leq & \int_X|h||f-g|d\mu\\ &\leq& ||f-g||_{L^2(X)}\\ &=& \sqrt{\int_Xf^2d\mu +\int_Xg^2d\mu - 2\int_Xfgd\mu}\\ &\leq& \sqrt{2 - 2\int_Xfgd\mu}\end{eqnarray}
I'm thinking C-S will not work here because it introduces a square root. Any hints would be greatly appreciated.
If $|a|\leq 1$ and $|b| \leq 1$ then $ab+|a-b| \leq 1$. [ For example, if $a \geq b$ this inequality says $a(1+b) \leq 1+b$ which is true]. Hence $|\int fhd\mu -\int gh d\mu| \leq \int |h||f-g| d\mu \leq \int |f-g| d\mu \leq 1-\int fg d\mu$.