Let $K/L/F$ be a field extension such that $L/F$ is purely inseparable. If $a\in K$ is separable over $F$ then $\min_L(a)=\min_F(a)$.
All I have so far is that since $a$ is separable then $min_F(a)=(x-a_1)...(x-a_n)$, if there exists an $i \in \{1,...,n\}$ such that $a_i \in L$ then it is very easy to prove that $min_L(a)=min_F(a)$. If not, I don't know how to proceed.
If $a\in K$ is separable over $F$, then $\min_L(a)$ divides $\min_F(a)$, hence $a$ is also separable over $L$. Write $$ \min_L(a) = (X-a_1)\dotsm (X-a_m) $$ for pairwise distinct elements $a_1 = a, a_2,\dotsc,a_m$ in an algebraic closure of $K$. Notice that these are all separable over $F$ (being roots of $\min_F(a)$). Expanding the above product, writing $$ \min_L(a) = \sum_{i=0}^m\lambda_iX^i\in L[X], $$ we see that $\lambda_i\in F(a_1,\dotsc,a_m)$ and hence they are separable over $F$. As the $\lambda_i$ also lie in $L$ and $L/F$ is purely inseparable, we conclude $\lambda_i\in F$ for all $i=0,\dotsc,m$. This shows $\min_L(a) = \min_F(a)$.