Prove that $\mu(\Omega)^2\leq\int f \,d\mu\int\frac{1}{f}\,d\mu$.
I don't know if that what I did is correct or if it will help to solve the problem, but here it is:
Using the Hölder inequality $\mu(\Omega)=\int_{\Omega} 1 \,d\mu=\int_{\Omega} |f.\frac{1}{f}|\,d\mu\leq (\int_{\Omega}|f|^1\, d\mu)^{\frac{1}{1}}(\int_{\Omega}|\frac{1}{f}|^1\, d\mu)^{\frac{1}{1}}= \int_{\Omega}f\, d\mu\int_{\Omega}\frac{1}{f}\, d\mu$.
I stopped here.
You are almost right. Just use the Cauchy-Schwarz inequality $$\mu(\Omega)=\int_{\Omega} 1 \,d\mu=\int_{\Omega} \left|f^{1/2}\frac{1}{f^{1/2}}\right|\,d\mu\leqslant \left(\int_{\Omega}|f|\, d\mu\right)^{\frac{1}{2}}\left(\int_{\Omega}\frac{1}{|f|}\, d\mu\right)^{\frac{1}{2}} $$