Which is correct $Var(\theta) \ge (F^{-1})_{ii} $ or $Var(\theta) \ge 1/(F_{ii}) $? With $\textbf{F}(\theta)$ denoting the Fisher matrix,  $\textbf{V}(\theta)$ the variance matrix, $\textbf{C}(\theta)$ the covariance matrix, and   parameter vector $\theta = [\theta_1, \theta_2, \dots, \theta_N]$. It is well known that the covariance matrix
$$\textbf{C}(\theta) \ge \textbf{F}^{-1}(\theta) ~~~~~~~(\star)$$
Also for a single parameter $\theta_i$ we know
$$\textbf{V}(\theta_i) \ge \frac{1}{\textbf{F}_{ii}}  ~~~~~~~(\star \star)$$
But from $(\star)$ $$\textbf{C}(\theta_i) \ge (\textbf{F}^{-1})_{ii}  ~~~~~~~(\star \star \star)$$
Given that from linear algebra $$(\textbf{F}^{-1})_{ii} \ge \frac{1}{\textbf{F}_{ii}},$$
how can we reconcile $(\star \star)$ and $(\star \star \star)$?

Foreword: I take it for granted that the preliminary conditions for the Cramér–Rao bound are satisfied. You always should validate it prior to usage.
The CRB for the general univariate case is
$$Var\left(\hat{\theta}\right)\geq \frac{\left(1+\frac{\partial}{\partial\theta}b\left(\hat{\theta}\right)\right)^2}{I(\theta)},$$
with $b\left(\hat{\theta}\right)$ being the bias function. If $\hat{\theta}$ is unbiased then $b\left(\hat{\theta}\right)=0$ and the numerator is 1.
Let $\psi(\theta)$ be an unbiased estimator of the vector $\theta$, the CRB of the multivariate unbiased case is
$$Cov\left(\psi(\theta)\right)\geq I(\theta)^{-1},$$
and for the estimator of $\theta_i$ you get
$$Var\left(\psi(\theta)_i\right)\geq \left( I(\theta)^{-1} \right)_{ii}.$$
The loose lower bound for reciprocal of diagonal is $\left( I(\theta)^{-1} \right)_{ii} \geq \left( I(\theta)_{ii}\right)^{-1} $ so overall you get
$$Var\left(\psi(\theta)_i\right)\geq \left( I(\theta)^{-1} \right)_{ii}\geq \left( I(\theta)_{ii}\right)^{-1}.$$
The univariate case has the slightly different same  bound because you have no problems of taking the reciprocal of a scalar. Overall, regarding the title of your questions - both bounds are correct, one of them is tighter.
