= Solution
For any chosen input distribution, the chain rule and nonnegativity of <conditional mutual information> give
$$
I(U;V,W)=I(U;V)+I(U;W\mid V)\geq I(U;V),
$$
and likewise $I(U;V,W)\geq I(U;W)$. Choosing an input distribution attaining either marginal <Channel capacity>[capacity] proves $K\geq\max\{K_1,K_2\}$.
For the upper bound, <conditional independence> implies $H(V,W\mid U)=H(V\mid U)+H(W\mid U)$, while <Shannon entropy>[entropy] subadditivity gives $H(V,W)\leq H(V)+H(W)$. Therefore
$$
\begin{aligned}
I(U;V,W)&=H(V,W)-H(V,W\mid U)\\
&\leq H(V)+H(W)-H(V\mid U)-H(W\mid U)\\
&=I(U;V)+I(U;W)\leq K_1+K_2.
\end{aligned}
$$
Maximizing over the input law gives
$$
\boxed{\max\{K_1,K_2\}\leq K\leq K_1+K_2.}
$$
More precisely, under the conditional-independence assumption,
$$
I(U;V,W)=I(U;V)+I(U;W)-I(V;W).
$$
The outputs may be dependent through the common input, and this final term measures the information they share rather than add independently.
This is <information from two conditionally independent observations>.
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