In addition to saying that $P$ and $Q$ are mutually absolutely continuous, you also need to require that expectations, defining Kullback divergence and $\chi^2$ divergence exist.
The inequality as stated does not hold. Indeed, consider $P$ a measure corresponding to $\operatorname{Beta}(2,1)$ random variable, and $Q$ to uniform random variable, i.e. $ \mathrm{d}P = 2 x \mathbf{1}_{(0,1)}(x) \mathrm{d} x \qquad \mathrm{d}Q = \mathbf{1}_{(0,1)}(x) \mathrm{d} x $ Then, it is easy to compute that $ K(P,Q) = \int_0^1 \log(2x) \cdot 2 x \mathrm{d} x = \log(2) - \frac{1}{2} \approx 0.19131472\ldots $ whereas $ \chi^2(Q,P) = \int_0^1 \left( 2x-1 \right)^2 \cdot 2 x \mathrm{d} x = \frac{1}{3} $ Clearly $K(P,Q) > \frac{1}{2} \chi^2(Q,P)$.
N.B. If $\operatorname{Beta}(2,2)$ is used for $Q$ and $\mathcal{U}(0,1)$ for $P$, then the integral defining $\chi^2(Q,P)$ is easily seen to diverge.