Among the many equivalent formulations of Leopoldt's conjecture, this one is probably the shortest: For any number field $K$, prime number $p$, finite set $S$ of primes of $K$ containing the primes above $p$, one has

Leopoldt's conjecture: $H^2(G_{K,S},\mathbb{Q}_p)=0$.

Here $G_{K,S}$ is as usual the Galois group of the maximal algebraic extension of $K$ un ramified outside $S$ and places at infinity, the $H^2$ is continuous cohomology.

Now, one of the most natural way to get a class in an $H^2$ is as a cup-product of two classes in an $H^1$. For example, if $\chi : G_{K,S} \rightarrow Q_p^\ast $ is a continuous character, then there is a cup-product map $$H^1(G_{K,S},\chi) \times H^1(G_{K,S},\chi^{-1}) \rightarrow H^2(G_{K,S},\mathbb{Q}_p),$$ which, according to Leopoldt's conjecture, should be zero.

Is it any easier to prove that the above morphism is zero than to prove Leopoldt's conjecture itself ?

I would also be interested to know the answer in special cases (of $K$, $\chi$, $p$) where Leopoldt's conjecture is not known.