On page 95 of Halmos' Naive Set Theory, we get the exercise
If $\{a_i\}$ and $\{b_i\}$ are families of cardinal numbers such that $a_i< b_i$, then $$\sum_i a_i<\prod_ib_i$$
I know that we get $\le$ easily (one can create an injection from the LHS to RHS). The real task is to prove a bijection cannot happen. I tried doing a diagonalization argument like in Cantor's theorem, but it didn't pan out. The arbitrary index $I$ makes it tough to do something like that. Even if we replace it with an ordinal, because it could be uncountable.
I'm thinking maybe one must use transfinite induction on the ordinals (on the index)?
Any tips/solutions?
Before the proof, I notice that the sum $\sum_{i\in I} a_i$ is identified to the disjoint union $\bigcup_{i\in I} a_i\times\{i\}$ for convenience.
You can prove the inequality is strict by diagonal argument. Remember that $|A|\le |B|$ if and only if there is an injection from $A$ to $B$. If every function from $A$ to $B$ is not onto, then $|A|<|B|$. From this, we should prove: if $f:\sum_{i\in I} a_i \to \prod_{i\in I} b_i$ is an injective function then there is a function $c\in \prod_{i\in I} b_i$ which is not an element of the range of $f$ (so $f$ should not be an bijection.)
Consider $f\upharpoonright a_i$, function $f$ whose domain is restricted from $a_i$. Since $a_i<b_i$, the cardinality of the range of $f\upharpoonright a_i$ is less than $b_i$. Also, the range of $f\upharpoonright a_i$ consists of functions $x$ from $I$ with $x(i)\in b_i$ for all $i\in I$. From above facts we get $$|\{x(i)\in b_i : x\in \text{range of }f\upharpoonright a_i\}|\le a_i<b_i$$ so we can choose $c_i\in b_i$, with $c_i\neq x(i)$ for all $x\in (\text{range of }f\upharpoonright a_i)$.
Let define $c\in\prod_{i\in I}b_i$ with $c(i)=c_i$. You can check that $c_i\in b_i$ for each $i$, and $c\neq f(x)$ for all $x\in \sum_i a_i$.