Not always. If $L/K$ is a Galois extension with non-cyclic Galois group, then there is no inert prime.
Indeed, if $p \subset O_K$ is inert in $L$, then the decomposition group $D_p \subset \mathrm{Gal}(L/K) = G$ is the whole Galois group $G$, and the inertia subgroup $I_p \subset G$ is trivial.
But it always holds that $D_p / I_p$ is isomorphic to the Galois group $H$ of the residual fields $O_L/P$ and $O_K/p$, which are finite (where $P$ is the given prime ideal of $O_L$ above $p$, in our case $P = pO_K$ is prime). In particular, $H$ is cyclic. Therefore $D_p / I_p \cong D_p = G$ is cyclic.
In other words, if $G$ is not cyclic, then there is no inert prime in $L/K$.
(And the converse is true also. If $G$ is cyclic then there are infinitely many inert primes in $L/K$. Indeed, if $\mathfrak p$ is a prime ideal in $\mathcal O_K$ that is unramified in $L$ and its Frobenius automorphism in $G$ is a generator of the cyclic group $G$, then the order of the Frobenius automorphism of $\mathfrak p$ is $|G|$ so the usual $efg = [L:K]$ equation for primes in a Galois extension turns into $1\cdot |G| g = |G|$, so $g = 1$ and thus $\mathfrak p\mathcal O_L$ is prime. The existence of infinitely many $\mathfrak p$ with any specific element of $G$ as its Frobenius automorphism follows from the Chebotarev density theorem.)
This fact has a nice application: the polynomial $x^4+1$ is irreducible over $\Bbb Q$ but is reducible mod $p$ for every prime integer $p$ (see here) !
What happens is that the splitting field of $x^4+1$ is $\Bbb Q(\zeta_8)/\Bbb Q$, which is a Galois and non-cyclic extension, so there are no inert primes in that extension. Indeed, if $p$ is inert, then $x^4+1$ would be irreducible (since $\Bbb Z[\zeta_8]$ is the ring of integers of $\Bbb Q(\zeta_8)$ — it is monogenic), see Neukirch's Algebraic number theory I.8.3.
