the channel does not maximize the mutual information, the source can select a distribution to maximize the mutual information and achieve a mutual information close to the capacity for a given channel. the best channel is an one to one mapping $f$ so $I(X,f(X))=I(X,X)=H(X)$. You can also use the data processing inequality for the Markov chain.

Sorry I dont have enough points to comment.

the channel does not maximize the mutual information, the source can select a distribution to maximize the mutual information and achieve a mutual information close to the capacity for a given channel. the best channel is an one to one mapping $f$ so $I(X,f(X))=I(X,X)=H(X)$. You can also use the data processing inequality for the Markov chain.

$I(V;U)\leq min \{ I(V;X,Y),I(U;X,Y)\}=min \{ I(V;Y),I(U;X)\}$

$I(V',U')\leq min \{ I(V';X,Y),I(U';X,Y)\}=min \{ I(V';Y),I(U';X)\}\leq min \{ I(V;Y),I(U;X)\} $

Sorry I don't have enough points to comment.