Problem: If $\displaystyle u \in F$ is separable over field $\displaystyle K$, where $\displaystyle F$ is an extension of $\displaystyle K$, and $\displaystyle v\in F$, then $\displaystyle K(u,v) = K(u+v)$. If $\displaystyle u, v \neq 0$, then $\displaystyle K(u,v) = K(uv)$.

It is clear to me that in characteristic 0 this proof is trivial. In characteristic $\displaystyle p$, with $\displaystyle p$ prime, one direction of the inclusion is obvious for each case, and that is $\displaystyle K(u+v), K(uv) \subset K(u,v)$. To get the other direction I was thinking of using the identity $\displaystyle (u + v)^{p^n} = u^{p^n} + v^{p^n}$, where $\displaystyle n$ is the power that makes $\displaystyle v^{p^n} \in K$. Therefore, we must have that $\displaystyle u^{p^n} \in K(u + v)$ by subtracting off $\displaystyle v^{p^n}$ (since it is an element of $\displaystyle K$). If $\displaystyle u$ is an element of $\displaystyle K(u+v)$, then we can subtract it off of the element $\displaystyle u + v \in K(u+v)$, and the reverse containment will follow. If $\displaystyle u \notin K(u+v)$, then it is purely inseparable over $\displaystyle K(u+v)$. Similar reasoning can be used to show $\displaystyle v$ is purely inseparable over $\displaystyle K(u+v)$ (otherwise $\displaystyle v$ would be an element of $\displaystyle K(u+v)$ and we'd be done), hence $\displaystyle K(u,v)$ is a purely inseparable extension of $\displaystyle K(u+v)$.

My issue is that I have no idea where to go with this; I'm hoping to find some kind of contradiction, but I am probably lacking somewhere in my understanding of what it means for an element to be separable. Any ideas? Is my approach even close to correct? I've been on this problem for 2 days, with no luck.