I am a bit rusty on scattering but I will try to give a satisfactory answer.
Real permittivity fluctuations acquire a tiny imaginary component from its own radiation reaction. This imaginary part results in a $\pi/2$ phase shift that ensures the Rayleigh field subtracts from the forward incident beam by exactly the amount of power that is re-radiated into the side lobes, so the wave is attenuated, not amplified. The bookkeeping is enforced by the optical theorem.
Start with a monochromatic dipole at the origin,
$$
\mathbf{p}(t) =\mathbf{p}_0 e^{-i\omega t}, \quad k= \frac{\omega}{c}.
$$
Expanding the retarted Green function about the origin gives (see Jackson Ch. 9, 16 and Nano-Optics by Novotny and Hecht Ch. 8-9)
$$
\mathbf{E}_{\text{self}}(0) = \frac{i k^3}{6\pi \epsilon_0}\mathbf{p}_0.
$$
In which the imaginary part (which by Euler's formula is $i=e^{i\pi/2}$, or a $\pi/2$ phase shift) leads the dipole by $+ 90^\circ$ in phase. This term is ultimately what protects energy conservation. For a tiny, loss-less dielectric blob the bare electrostatic polarizability $\alpha_0$ is real. Because the local field is External + Self, the induced dipole must satisfy
$$
\mathbf{p}_0 = \alpha_0 \left(\mathbf{E}_{\text{inc}} + \frac{i k^3}{6\pi \epsilon_0}\mathbf{p}_0 \right).
$$
Solve for $\mathbf{p}_0$ to get
$$
\alpha_{\text{eff}} = \frac{\alpha_0}{1-\frac{ik^3 \alpha_0}{2\pi \epsilon_0}}.
$$
For Rayleigh particles $k^3 \alpha_0 << 1$ we may expand
$$
\alpha_{\text{eff}} \approx \alpha_0 + i\frac{\alpha_0^2 k^3}{6\pi \epsilon_0}.
$$
So even the most harmless real polarizability picks up a small imaginary component whose magnitude is $\propto k^3$. Say we align the incident plane wave with $+\hat{z}$ and its electric field with $\hat{x}$. The Rayleigh dipole radiates
$$
\mathbf{E}_{\text{sc}}^0(z) = \frac{k^2 \alpha_{\text{eff}}}{4\pi \epsilon_0 z}\mathbf{E}_{\text{inc}} e^{ikz}.
$$
Because the spherical factor $e^{ikz}/z$ has the same phase as the incident wave along $+z$, any phase difference must come from $\alpha_{\text{eff}}$. Split the latter into its real and imaginary parts:
$$
\mathbf{E}_{\text{sc}}^0 = \frac{k^2 \alpha_0}{4\pi \epsilon_0 z}\mathbf{E}_{\text{inc}} + i \frac{k^2\alpha_0^2 k^3}{24 \pi^2 \epsilon_0^2 z}\mathbf{E}_{\text{inc}}.
$$
The first term is in phase and merely tweaks the plane wave's phase velocity (i.e. gives a refractive-index shift). The second term is $+\pi/2$ out of phase and therefore capable of subtracting power from the beam. Define $f(0) = k^2 \alpha_{\text{eff}}/(4\pi \epsilon_0)$. The optical theorem states
$$
\sigma_{\text{ext}} = \frac{4\pi}{k}\text{Im}f(0) = \int |f(\theta)|^2 \ d\Omega = \sigma_{\text{sca}}.
$$
The LHS uses only the imaginary part of the forward amplitude. The RHS is the total Rayleigh power radiated into all angles. Thus, the tiny $+\pi/2$ component of $\alpha_{\text{eff}}$ removes exactly the power that reappears as scattered light. Another way of showing this is calculating the power removed from the beam by interference, where we integrate the cross Poynting term over a remote plane $z=z_0$. Then
$$
P_{\text{lost}} = I_0 \frac{4\pi}{k}\text{Im}f(0) = I_0 \frac{k^4 \alpha_0^2}{6\pi \epsilon_0^2}.
$$
The total (angle-integrated) scattered power is
$$
P_{\text{sca}} = I_0 \frac{k^4|\alpha_\text{eff}|^2}{6\pi \epsilon_0^2} = I_0 \frac{k^4 \alpha_0^2}{6\pi \epsilon_0^2}
$$
after we drop the $O(k^6)$ correction term coming from the $\text{Im}\ \alpha_{\text{eff}}<<\alpha_0$. We then see that the power lost is the power scattered, so no amplification.
