Given a function $f(x,t)$ and an average measure defined as $\langle \bullet \rangle \equiv \int_\infty dx \, g(x) f(x)$, where $g(x)$ is time-independent and normalized such that $\int g^2(x)dx=1$, I want to show that \begin{equation} \left \langle \frac{d \cos^2f}{dt} \right \rangle^2 \leq ? \left \langle \cos^2f \right \rangle \left \langle \sin^2f \right \rangle \left \langle \left( \frac{d \cos^2f}{dt} \right)^2\right \rangle \end{equation}
The function $f$ is well behaved, and we can define a bound to its time-evolution. Hence the problem comes to solving \begin{equation}\label{question} \left \langle \cos f \sin f \right \rangle^2 \leq ? \left \langle \cos^2f \right \rangle \left \langle \sin^2f \right \rangle \left \langle \cos^2f \sin^2 f \right \rangle \quad \quad (1) \end{equation}
Using Cauchy-Schwarz and the Gram's inequality, I have so far \begin{eqnarray} &&\left \langle \cos^2 f \sin^2 f \right \rangle \left \langle \cos f \sin f \right \rangle^2 \\ &\leq& \left \langle \cos^2 f \sin^2 f \right \rangle \left \langle \cos f \sin f \right \rangle^2 + \left \langle \left( \cos f \langle \cos f \sin^2 f \rangle - \sin f \langle \cos^2 f \sin f \rangle\right)^2 \right \rangle \\ &\leq &\left \langle \cos^2f \right \rangle \left \langle \sin^2f \right \rangle \left \langle \cos^2f \sin^2 f \right \rangle \end{eqnarray}
I need a tighter bound to prove the sign in Eq. (1).
Any suggestion is welcomed!