unexplained inequalities from Niven's 1971 proof $\sum 1/p$ diverges

He explains his reasoning with respect to the first inequality in the paragraph above the inequality. Since each integer can be uniquely expressed as the product of a square-free integer and a square, each term on the right-hand sum occurs as a product of a single term from the square-free sum and a single term from the square sum. The inequality isn't strict because as $k$ and $j$ approach $n$, their product may exceed $n$, yielding some additional terms in the sum.

Similarly, in the second inequality, if you simply expand the product on the left, you will get all of the terms in the right-hand sum, and possibly some additional terms that you can throw away.

Each $m<n$ can be uniquely expressed as $m=kj^2$ where $k$ is a square-free integer, therefore $$ \sum_{m<n}\frac{1}{m}=\sum_{\substack{k,j \\ kj<n}}\frac{1}{kj^2}\leqslant\sum_{k,j<n}\frac{1}{kj^2}=\left(\sum_{k<n}'\frac{1}{k}\right)\left(\sum_{j<n}\frac{1}{j^2}\right) $$ where $k$ is a square free integer in the second and the third sum. As for the last inequality, let $p_1,\ldots,p_r$ be the prime numbers $<n$, then developping the product gives $$ \prod_{p<n}\left(1+\frac{1}{p}\right)=\sum_{I\subset[\![1,r]\!]}\prod_{i\in I}\frac{1}{p_i}$$ Each square-free $k<n$ has their prime divisors among $p_1,\ldots,p_r$ and thus is of the form $\prod_{i\in I}p_i$ where $I\subset[\![1,r]\!]$, therefore $$\prod_{p<n}\left(1+\frac{1}{p}\right)\geqslant\sum_{k<n}'\frac{1}{k} $$

When you multiply out the product

$$\left(\sum_{k<n}'\frac1k\right)\left(\sum_{j<n}\frac1{j^2}\right)\,,\tag{1}$$

you get the sum of all possible terms of the form $\frac1{kj^2}$, where $k$ is a square-free integer less than $n$, and $j$ is any integer less than $n$. Every positive integer $m<n$ can be expressed as the product of such a $k$ and $j^2$, so these product terms include every fraction $\frac1m$ for $m<n$. In general, however, they will also include some fractions $\frac1m$ with $m\ge n$: if $n-1$ is square-free, for instance, we can take $k=j=n-1$ and get $\frac1{(n-1)^3}$. Thus, every term of $\sum_{m<n}\frac1m$ appears in $(1)$ after the latter is multiplied out, and $(1)$ generally contains extra terms as well, so

$$\left(\sum_{k<n}'\frac1k\right)\left(\sum_{j<n}\frac1{j^2}\right)\ge\sum_{m<n}\frac1m\,.$$

For the second inequality let $P=\{p_1,\ldots,p_\ell\}$ be the set of primes less than $n$. When the product

$$\prod_{p<n}\left(1+\frac1p\right)=\prod_{i=1}^\ell\left(1+\frac1{p_i}\right)\tag{2}$$

is multiplied out, there is one term for each possible choice of one factor, $1$ or $\frac1{p_i}$, from each of the $\ell$ factors. Specifically, if $S\subseteq P$, one of the terms of the product is

$$\frac1{\prod_{i\in S}p_i}\,,\tag{3}$$

obtained by choosing $1$ when $i\notin S$ and $\frac1{p_i}$ when $i\in S$.

Every square-free integer less than $n$ is a product of distinct primes less than $n$, so every term of $\sum\limits_{k<n}'\frac1k$ is one of the terms $(3)$. However, there may also be some $S\subseteq P$ such that $\prod_{i\in S}p_i\ge n$, so the terms of $(2)$ after it’s multiplied out include all of the terms of $\sum\limits_{k<n}'\frac1k$ and possible some others as well. Thus,

$$\prod_{p<n}\left(1+\frac1p\right)\ge\sum_{k<n}'\frac1k\,.$$