= Solution
This is <König theorem for cardinal numbers>. Since $\kappa_i<\lambda_i$, map $(i,\alpha)$ in the disjoint union to the element $s_{i,\alpha}$ of the product defined by
$$
s_{i,\alpha}(j)=
\begin{cases}
\alpha+1,&j=i,\\
0,&j\ne i.
\end{cases}
$$
Here $\alpha<\kappa_i$ implies $\alpha+1<\lambda_i$. The unique nonzero coordinate and its value recover $(i,\alpha)$, so this is an injection
$$
\bigsqcup_{i\in I}\kappa_i\longrightarrow
\prod_{i\in I}\lambda_i.
$$
For strictness, suppose $F$ mapped the disjoint union onto the product. For each $i$, the set
$$
A_i=\{F(i,\alpha)(i):\alpha<\kappa_i\}
$$
has cardinality at most $\kappa_i$, so it cannot exhaust $\lambda_i$. Choose $b_i\in\lambda_i\setminus A_i$. Then $b=(b_i)_{i\in I}$ differs from $F(i,\alpha)$ in coordinate $i$ for every $(i,\alpha)$, contradicting surjectivity. By <Cantor-Schröder-Bernstein theorem>, an injection in the reverse direction would combine with the displayed injection to give a bijection and hence a surjection. Therefore
$$
\boxed{\sum_{i\in I}\kappa_i<\prod_{i\in I}\lambda_i.}
$$
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