The mean of the Beta distribution is
$$\mu = \frac {\alpha}{\alpha + \beta}$$
We want to see whether restricting the permissible range of $\mu$, will guarantee that we will have either $\{\alpha \geq 1, \beta >0\}$, OR $\{\alpha >0, \beta \geq 1\}$.
Treating the mean as a function of the parameters we obtain
$$\frac {\partial \mu}{\partial \alpha} > 0, \;\; \frac {\partial \mu}{\partial \beta} <0$$
So it is monotonically increasing in $\alpha$ and monotonically decreasing in $\beta$.
So
$$\min \mu = 1/(1+\beta)\implies \{\alpha \geq 1, \beta >0\} \tag {1}$$
but the situation $\{\alpha >0, \beta \geq 1\}$ permits all possible values of $\mu$ (in $(0,1)$).
In other words by restricting the mean to lie in the interval $[1/(1+\beta),\, 1)$ we can guarantee that we will have $\{\alpha \geq 1, \beta >0\}$. But there is no restriction on the mean that will guarantee us that $\{\alpha >0, \beta \geq 1\}$.
So we should turn to the variance, which is
$$\sigma^2 = \frac {\alpha \beta}{(\alpha + \beta)^2(\alpha + \beta +1)}$$
It is not difficult to determine that no restriction on the range of the variance can guarantee that we will have $\{\alpha >0, \beta \geq 1\}$.
So the answer to the question is that the best we can do is to guarantee what relation $(1)$ reflects, by restricting the value of the mean from below.