Let $X_1,\ldots, X_n \sim f(x\mid \theta)=\frac{x}{\theta}e^{-x^2/(2\theta)}, x > 0$ independently. $\theta > 0$ is unknown.
How can I show that $\sum X_i$ is not a sufficient statistic for $\theta$?
I found from the likelihood ratio that
$$\frac{f(\mathbf{x}\mid \theta)}{f(\mathbf{y}\mid \theta)}=\prod^n_{i=1}\left(\frac{x_i}{y_i}\right)\frac{e^{-\frac{1}{2\theta}\sum x_i^2}}{e^{-\frac{1}{2\theta}\sum y_i^2}}$$
Since we need this likelihood ratio to not be a function of $\theta$, $\sum x_i^2=\sum y_i^2$. Thus, $T(\mathbf{X})=\sum x_i^2$ is our minimal sufficient statistic. (Is this logic correct?)
How can I show that $\sum x_i$ cannot be a sufficient statistic since $T(\mathbf{X})=\sum x_i^2$ cannot be written as a function of $\sum x_i$?