I am trying to show that squeezed states are non classical by showing that the Glauber-Sudarshan $P$ function takes negative values. A squeezed state is one for which one of the quadratures $\Delta X_1 < \frac{1}{2}$. In Introductory quantum optics by Gerry and Knight on page 152 they show that for the quadrature $X_1$ we have
$$ \langle (\Delta X_1)^2 \rangle = \frac{1}{4} \bigg\{1+ \int P(\alpha)[(\alpha +\alpha^*)-(\langle a \rangle- \langle a^\dagger \rangle)]^2d^2\alpha\bigg\}.$$
I am trying to derive this from the fact that for an observable $A$ and a density operator $\rho$,$\langle A \rangle = \mathrm{tr}(A \rho)$. We can express the density operator in terms of the Glauber-Sudarshan $P$ function as
$$ \rho = \int d^2 \alpha P(\alpha)|\alpha \rangle \langle \alpha |$$
where $ \{|\alpha \rangle \}$ are coherent states. So working through the algebra I find
\begin{equation} \begin{split} \langle (\Delta X_1)^2\rangle & = \mathrm{tr}\bigg[ \rho (\Delta X_1)^2 \bigg] \\ & = \mathrm{tr} \bigg[ \int \mathrm{d}^2 \alpha P(\alpha) |\alpha \rangle \langle \alpha | (\Delta X_1)^2 \bigg] \\ & = \int \mathrm{d}^2 \alpha P(\alpha) \langle \alpha | (\Delta X_1)^2 |\alpha \rangle \\ &= \frac{1}{4} \int \mathrm{d}^2 \alpha P(\alpha) \end{split} \end{equation} where I used the fact that $\langle \alpha | (\Delta X_1)^2 |\alpha \rangle = \frac{1}{4}$ for a coherent state $|\alpha \rangle$. The density matrix has unit trace so $\int \mathrm{d}^2 \alpha P(\alpha)=1$ and I get $(\Delta X_1)^2 =\frac{1}{4}$ which does not agree with Gerry and Knight. Where have I gone wrong?