Caldeira-Leggett Dissipation: frequency shift due to bath coupling

If you look at the equation of motions of $Q$ and the $q_i$, and solve them for the bath, then inject this in the equation for $Q$, you will see that the frequency of $Q$ is shifted by the amount quoted in the question.

If you want to define the true frequency of the system (in presence of the bath) by $\Omega^2$, you have to put this shift in the Lagrangian.

PS: because the system is quadratic, solving the classical equations of motion is equivalent to the quantum problem for the purpose of the question.

To expand on Adam's answer, let's see the effect of a static $Q$ on the system without the $\Delta \Omega^2$. The Lagrangian becomes $$L = -\frac{1}{2}\Omega^2Q^2 - Q\sum_i f_iq_i + \sum_i \frac{1}{2}(\dot{q}_i^2 - \omega_i^2q_i^2)$$ The equation of motion for each of the $q_i$ is now $$\frac{\partial L}{\partial q_i} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q_i}} \implies -Qf_i -\omega_i^2q_i = \ddot{q}_i$$ To recover a harmonic oscillator we shift the coordinates by defining $q_i' =q_i + Q\frac{f_i}{\omega_i^2}$, and now the EOM is the familiar $\ddot{q'_i} = -\omega_i^2q_i'$. We can immediately identify the static shift as the quantity we subtracted out of the $q_i$, i.e. $$\Delta q_i = -Q\frac{f_i}{\omega_i^2}$$ Multiply both sides by $f_i$ and you recover the equation you want.

Why does this effect the frequency of the macro-oscillator $Q$? Intuitively, you might think that all of the environment oscillators $q_i$ are "dragging" on $Q$. Quantitatively, replace all of the $q_i$ in the Lagrangian with the $\Delta q_i$ above and you'll find the EOM for $Q$ is now $$\ddot Q = -\left(\Omega^2 - \sum_i\frac{f_i^2}{\omega_i^2} \right)Q$$ which gives you the frequency shift $\Delta \Omega^2$.

Note: Much of this is from Stone & Goldbart's excellent textbook Mathematics for Physics.