Infinitude of primes which are inert in one extension and completely split in other for extensions of prime degree. I want to prove the following:

Let $L_1, \cdots , L_n$ be extensions of number field $K$ of prime degree $p$ such that each $L_i$ is disjoint from the compositum of $L_j$s for $ j\neq i $. Then there are infinitely many primes of $K$ that are inert in $L_1$ and split completely in $L_i$ for $i \geq 2 $.

I need to prove this to prove the second inequality of Global Class Field Theory (so I do not want to use the Chebotarev density theorem because that can give me the second inequality directly instead).
I have a couple of facts, consequences of the first inequality, at my disposal, which I think might be useful here.


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*In a non-trivial finite abelian extension of $K$, infinitely many primes of $K$ do not split completely.

*In a cyclic prime power degree extension of $K$, infinitely many primes remain inert.


I first try to prove the theorem in the case where we just have $L_1$ and $L_2$, hoping that the proof will generalize easily.
In $L_1L_2$ there are infinitely many primes which do not split completely. There are infinitely many of these for which the prime lying below in $L_1$ is inert, let $S$ be set of such primes. We should show that infinitely many of primes $S$ when contracted to $L_2$ are completely split. I am stuck here.
 A: Right... I’m not entirely sure how to generalize this (and I’m not 100 percent sure it works), but in the case $n=2,L_i/K$ Galois, so that $L_1L_2$ is the tensor product and is Galois of group $\mathbb{F}_p^2$ (it’s of order $p^2$ and consider its subgroups).
Let $\mathfrak{q}$ be a prime of $L_2$ (above $\mathfrak{p}$ of $K$) which is inert in $L_1L_2$. There are infinitely many such $\mathfrak{p}$ that are unramified in $L_1L_2$.
Assume $\mathfrak{p}$ inert in $L_2$, then $f_{L_1L_2/K,\mathfrak{p}}=p^2$ by multiplicativity. It implies in particular that $\mathfrak{p}$ is inert in $L_1$. But then $L_1 \otimes K_{\mathfrak{p}}=L_2\otimes K_{\mathfrak{p}}$ (they’re unramified of same degree over $K_{\mathfrak{p}}$ – because Galois and prime) so $L_1L_2$ splits, a contradiction.
So such a $\mathfrak{p}$ splits in $L_2$. Assume that it splits in $L_1$, too. Then $(L_1L_2) \otimes K_{\mathfrak{p}}=L_1 \otimes (L_2 \otimes K_{\mathfrak{p}} \cong L_1 \otimes K_{\mathfrak{p}}^p=(L_1 \otimes K_{\mathfrak{p}})^p=K_{\mathfrak{p}}^{p^2}$ thus $\mathfrak{p}$ splits totally in $L_1L_2$, a contradiction.
So the primes $\mathfrak{p}$ of $K$ unramified in $L_1L_2$ such that there are primes $\mathfrak{q}$ of $L_2$ above them that are inert in $L_1L_2$ are infinitely many solutions to your question.
I think a similar argument (with $L_2\ldots L_n$ instead of $L_2$) works in general if each $L_i/K$ is Galois (with some minor adaptations) but I’m not sure.
