lagrangian formalism - Caldeira-Leggett Dissipation: frequency shift due to bath coupling

I am trying to understand the Caldeira-Leggett model. It considers the Lagrangian

$$L = \frac{1}{2} \left(\dot{Q}^2 - \left(\Omega^2-\Delta \Omega^2\right)Q^2\right) - Q \sum_{i} f_iq_i + \sum_{i}\frac{1}{2} \left(\dot{q}^2 - \omega_i^2q^2\right)$$

where $Q$ is the generalised coordinate of the macro variable (an oscillator with natural frequency $\Omega$), $q_i$ are variables related to an array of harmonic oscillators each with natural frequency $\omega_i$.The first term describes the potentail and kinetic energies related to the macro degree of freedom, the second term describes the coupling using constants $f_i$, the third again describes potential and kinetic energies of the array of oscillators,

$$\Delta \Omega^2 = -\sum_i \left( \frac{f_i}{\omega_i} \right)^2$$

is the ad hoc term my first question relates to. The explanation I found goes

the quantity is inserted to cancel the frequency shift $$\Omega^2 \to \Omega^2 - \sum_{i} \left(\frac{f_i}{\omega_i}\right)^2$$ [...] the shift arises because a static Q displaces the bath oscillators so that $$f_i q_i = - \left(f_i^2 / \omega_i^2\right)Q$$ Substituing these values for the $f_i q_i$ into the potential terms shows that the effective potentail seen by $Q$ would have a "shifted" frequency.

I regretfully do not get it. I tried to get to the equation

$$f_i q_i = - \left(f_i^2 / \omega_i^2\right)Q$$

by considering the equation of motion in equilibrium, without success. Why would $\Omega$ be affected by the coupling? Any hint would be so appreciated.