[proofplan] We identify cardinals with their initial ordinal representatives and realize the cardinal sum as a disjoint union and the cardinal product as a set of choice functions. First we construct an injection from the disjoint union into the product, so the sum is at most the product. Then we prove, by a diagonal argument, that no map from the disjoint union onto the product can be surjective. This excludes an injection in the opposite direction and therefore gives the strict cardinal inequality. [/proofplan] [step:Handle the empty index set separately] If $I = \varnothing$, then the cardinal sum is $0$ and the cardinal product is $1$, since the empty product is represented by the singleton set containing the empty function. Hence \begin{align*} \sum_{i \in \varnothing} \kappa_i = 0 < 1 = \prod_{i \in \varnothing} \lambda_i. \end{align*} For the rest of the proof, assume $I \neq \varnothing$. [/step] [step:Represent the sum and product by explicit sets] For each $i \in I$, regard $\kappa_i$ and $\lambda_i$ as initial ordinals. Define the disjoint-union representative of the cardinal sum by \begin{align*} S := \{(i,\alpha) : i \in I \text{ and } \alpha \in \kappa_i\}. \end{align*} Define the product representative by \begin{align*} P := \left\{ f: I \to \bigcup_{i \in I} \lambda_i : f(i) \in \lambda_i \text{ for every } i \in I \right\}. \end{align*} Then $|S| = \sum_{i \in I}\kappa_i$ and $|P| = \prod_{i \in I}\lambda_i$ by the definitions of cardinal sum and cardinal product. [/step] [step:Inject the disjoint union into the product] Since $\kappa_i < \lambda_i$ for every $i \in I$, each $\lambda_i$ is nonzero, so $0 \in \lambda_i$ for every $i \in I$. Define \begin{align*} E: S &\to P \\ (i,\alpha) &\mapsto E(i,\alpha), \end{align*} where $E(i,\alpha): I \to \bigcup_{j \in I}\lambda_j$ is the function \begin{align*} E(i,\alpha)(j) := \begin{cases} \alpha + 1, & j = i,\\ 0, & j \neq i. \end{cases} \end{align*} This is well-defined: if $\alpha \in \kappa_i$, then $\alpha + 1 \leq \kappa_i < \lambda_i$, hence $\alpha + 1 \in \lambda_i$, and $0 \in \lambda_j$ for $j \neq i$. The map $E$ is injective. Indeed, if $E(i,\alpha) = E(j,\beta)$ and $i \neq j$, then evaluating at $i$ gives \begin{align*} \alpha + 1 = E(i,\alpha)(i) = E(j,\beta)(i) = 0, \end{align*} which is impossible for an ordinal successor. Hence $i = j$. Evaluating at this common index gives $\alpha + 1 = \beta + 1$, and therefore $\alpha = \beta$. Thus $|S| \leq |P|$. [guided] The purpose of this step is to prove the non-strict inequality \begin{align*} \sum_{i \in I}\kappa_i \leq \prod_{i \in I}\lambda_i. \end{align*} Since we are using explicit representatives, it is enough to build an injection $S \to P$. Because $\kappa_i < \lambda_i$, the ordinal $\lambda_i$ is nonzero for every $i \in I$, so every coordinate set $\lambda_i$ contains $0$. We encode an element $(i,\alpha)$ of the disjoint union by a function that is zero everywhere except at the $i$th coordinate. At the $i$th coordinate we use $\alpha + 1$ rather than $\alpha$, so that different summands cannot collide at the all-zero function. Define \begin{align*} E: S &\to P \\ (i,\alpha) &\mapsto E(i,\alpha), \end{align*} where \begin{align*} E(i,\alpha)(j) := \begin{cases} \alpha + 1, & j = i,\\ 0, & j \neq i. \end{cases} \end{align*} This function belongs to $P$: for $j \neq i$ we have $0 \in \lambda_j$, while for $j = i$ the inequality $\alpha \in \kappa_i$ implies $\alpha + 1 \leq \kappa_i < \lambda_i$, so $\alpha + 1 \in \lambda_i$. Now suppose $E(i,\alpha) = E(j,\beta)$. If $i \neq j$, then evaluating both functions at the coordinate $i$ gives \begin{align*} \alpha + 1 = E(i,\alpha)(i) = E(j,\beta)(i) = 0, \end{align*} contradicting the fact that $\alpha + 1$ is a successor ordinal and hence nonzero. Therefore $i = j$. Evaluating at this same index gives $\alpha + 1 = \beta + 1$, and cancellation of ordinal successors gives $\alpha = \beta$. Thus $E$ is injective, and so $|S| \leq |P|$. [/guided] [/step] [step:Diagonalize against every map from the sum to the product] Let \begin{align*} F: S \to P \end{align*} be any map. For each $i \in I$, define the $i$th coordinate restriction \begin{align*} F_i: \kappa_i &\to \lambda_i \\ \alpha &\mapsto F(i,\alpha)(i). \end{align*} Since $\kappa_i < \lambda_i$, the map $F_i$ cannot be surjective. Therefore the set $\lambda_i \setminus F_i[\kappa_i]$ is nonempty. Because $\lambda_i$ is an ordinal, it is well-ordered, so define \begin{align*} a_i := \min(\lambda_i \setminus F_i[\kappa_i]) \end{align*} for each $i \in I$. Define \begin{align*} a: I &\to \bigcup_{i \in I}\lambda_i \\ i &\mapsto a_i. \end{align*} Then $a \in P$. We claim that $a \notin F[S]$. If $a = F(i_0,\alpha_0)$ for some $(i_0,\alpha_0) \in S$, then evaluating at $i_0$ gives \begin{align*} a_{i_0} = a(i_0) = F(i_0,\alpha_0)(i_0) = F_{i_0}(\alpha_0). \end{align*} This contradicts the definition of $a_{i_0}$ as an element of $\lambda_{i_0} \setminus F_{i_0}[\kappa_{i_0}]$. Hence no map $F:S \to P$ is surjective. [guided] We now prove the strict part by showing that every map from the disjoint union $S$ into the product $P$ misses at least one point of $P$. Fix an arbitrary map \begin{align*} F: S \to P. \end{align*} For each index $i \in I$, we inspect only the $i$th summand of $S$ and only the $i$th coordinate of the output. This gives a map \begin{align*} F_i: \kappa_i &\to \lambda_i \\ \alpha &\mapsto F(i,\alpha)(i). \end{align*} The hypothesis $\kappa_i < \lambda_i$ means that no function from a set of cardinality $\kappa_i$ can cover a set of cardinality $\lambda_i$. Therefore $F_i$ is not surjective, so there exists at least one element of $\lambda_i$ not hit by $F_i$. To choose the missed value canonically, use the fact that $\lambda_i$ is an ordinal and hence well-ordered. Define \begin{align*} a_i := \min(\lambda_i \setminus F_i[\kappa_i]). \end{align*} This gives one missed value in each coordinate. Now assemble these missed values into a single product element: \begin{align*} a: I &\to \bigcup_{i \in I}\lambda_i \\ i &\mapsto a_i. \end{align*} Since $a_i \in \lambda_i$ for every $i \in I$, the function $a$ belongs to $P$. We show that $a$ is not in the range of $F$. Suppose, toward a contradiction, that $a = F(i_0,\alpha_0)$ for some $(i_0,\alpha_0) \in S$. Then the $i_0$th coordinate of this equality says \begin{align*} a_{i_0} = a(i_0) = F(i_0,\alpha_0)(i_0) = F_{i_0}(\alpha_0). \end{align*} But $a_{i_0}$ was chosen outside the image $F_{i_0}[\kappa_{i_0}]$, while the displayed equality places it inside that image. This contradiction proves that $a \notin F[S]$. Since $F$ was arbitrary, no map $S \to P$ is surjective. [/guided] [/step] [step:Convert the diagonal obstruction into strict cardinal inequality] We have shown that $|S| \leq |P|$. It remains to rule out $|P| \leq |S|$. Suppose, toward a contradiction, that there is an injection \begin{align*} G: P \to S. \end{align*} Let \begin{align*} z: I &\to \bigcup_{i \in I}\lambda_i \\ i &\mapsto 0 \end{align*} be the zero element of $P$. Define \begin{align*} H: S &\to P \end{align*} by setting $H(s)$ equal to the unique $p \in P$ such that $G(p)=s$ when $s \in G[P]$, and setting $H(s)=z$ when $s \notin G[P]$. Since $G$ is injective, this definition is unambiguous. Moreover, for every $p \in P$, \begin{align*} H(G(p)) = p, \end{align*} so $H$ is surjective. This contradicts the previous step, which showed that no map $S \to P$ is surjective. Hence $|P| \nleq |S|$. Combining $|S| \leq |P|$ with $|P| \nleq |S|$, we obtain \begin{align*} |S| < |P|. \end{align*} Using $|S| = \sum_{i \in I}\kappa_i$ and $|P| = \prod_{i \in I}\lambda_i$, this is exactly \begin{align*} \sum_{i \in I}\kappa_i < \prod_{i \in I}\lambda_i. \end{align*} [/step]