Here is a small observation, generalizing Lucia's comment.

**Proposition.** If $A$ is a set of real numbers with minimal distance at least $1$, then $$|A+AA| \geq \frac{|A|(|A|-1)}{2}\geq |A+A|-|A|.$$

**Proof.** Let $r_m>\dots>r_1>0$ be the positive elements of $A$. Then the subsets $r_i+r_m A$ of $A+AA$ are pairwise disjoint, because $r_i+r_ma=r_j+r_ma'$ $(i\neq j)$ would imply
$$ r_m\leq |r_m(a-a')|=|r_i-r_j|<r_m.$$
Hence $|A+AA| \geq m|A|$. Similarly, let $s_n<\dots<s_1<0$ be the negative elements of $A$. Then the subsets $s_i+s_n A$ of $A+AA$ are pairwise disjoint, because $s_i+s_na=s_j+s_na'$ $(i\neq j)$ would imply
$$ |s_n|\leq |s_n(a-a')|=|s_i-s_j|<|s_n|.$$
Hence $|A+AA| \geq n|A|$. It follows that
$$ |A+AA| \geq\max(m,n)|A|\geq\frac{m+n}{2}|A|\geq\frac{|A|-1}{2}|A|\geq |A+A|-|A|.$$

**Remark.** If $m\neq n$ and $0\not\in A$, then the last display improves to
$$|A+AA| \geq\max(m,n)|A|\geq\frac{m+n+1}{2}|A|=\frac{|A|+1}{2}|A|\geq |A+A|.$$