Let $R$ be a commutative ring. If $M$ and $N$ are finite length $R$-modules, then $M\otimes_R N$ has finite length, and $l(M\otimes_R N) \le l(M)l(N)$.
I know the question has been posted previously here, but only half of it was answered and I do not understand the argument about the inequality.
I have tried to explicitly construct a composition series and use exact sequences, but in using the latter I can't see what modules to use in the sequences.
Let $\{0\}=N_0<N_1<\cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^n\to N$, and tensorizing by $M$ we get $M\otimes_R R^n\to M\otimes_RN\to 0$. But $M\otimes_R R^n\simeq M^n$, so $M\otimes_R R^n$ has finite length which is greater or equal than the length of $M\otimes_RN$. Since the length of $M^n$ is $\ell(M)\ell(N)$, we are done.