The question reads : Find the maximum likelihood estimator for theta based on the sample of size n from a distribution with density $$f(x|\theta)=\frac{2\theta^2}{x^3};x>\theta.$$ According to my calculations it is observed that the maximum is not attained at a critical point, however I thought it fit to assume theta to be more than zero but not very much sure whether my conclusion will be nice that way.
$$L(\theta)=2n\ln(\theta)+n\ln2-3(\sum \ln(x))$$
$L'(\theta)=\frac{2}{\theta}$ which maximizes the the pdf but I am finding it difficult to conclude from the derivative with respect to theta.
Thank you in advance.
Elaborating on on @Did's comment, we find that $$\frac\partial{\partial \theta}\ell(\theta)=\frac{2n}\theta>0$$ for all $\theta$. Hence the MLE is $$\operatorname{argmax}_\theta \ell(\theta) = \max\{\theta : x_i \geqslant \theta\; \forall i\}=\min_i x_i. $$