I have to show that for $\psi$ and $\phi$ two positive and measurable functions:
$$\int_X \phi\psi \, d\mu \le \sqrt{\int_X \phi^2 \, d\mu} \sqrt{\int_X \psi^2 \, d\mu}$$
I know that for $f,g$ two measurable functions it holds that:
$$\int_X fg \, d\mu \le \frac{1}{2} \left(\int_X f^2 \, d\mu + \int_X g^2 \, d\mu\right)$$
I have set $f=\frac{ \phi}{\sqrt{\int \phi^2}}$ and $g=\frac{\psi }{ {\sqrt{\int \psi^2 d\mu}}}$ and then tried to plug them into the equation $2fg \le f^2 +g^2$
which lead to the following expression: $$\frac{\phi\psi}{\sqrt{\int \phi^2}\sqrt{\int \psi^2}} \le \frac{\phi^2}{\int \phi^2} + \frac{\psi^2}{\int \psi^2}$$
$$\phi\psi \le \sqrt{\int \phi^2} \sqrt{\int \psi^2} \left(\frac{\phi^2 \int \psi^2 + \psi^2 \int \phi^2}{\int \phi^2 \int \psi^2}\right) = \sqrt{\int \phi^2}\sqrt{\int \psi^2}(\phi^2+\psi^2)$$
I'm not sure if I have taken the wrong approach or if there is a way I can continue from this point? Thanks for any help! :)
The $f$ and $g$ that you defined have the property that $\int f^2=\int g^2=1$. So you have $$ \int fg\,d\mu\leq\frac12(1+1)=1. $$ So $$ \int_X\frac{\phi}{\sqrt{\int_X\phi^2\,d\mu}}\,\frac{\psi}{\sqrt{\int_X\psi^2\,d\mu}}\,d\mu\leq1, $$ which is what you want to prove.