A new bound for $A(A+A)$ for large sets

For $p$ being a large prime number, and $A\subset {\mathbb{F}}_p$ we prove the following: $(i)$ If $A(A+A)$ does not cover all nonzero residues in ${\mathbb{F}}_p$, then $\mathrm{\mid }A\mathrm{\mid }<p\mathrm{/}8+o(p)$. $(ii)$ If $A$ is both sum-free and satisfies $A=A^{\ast }$, then $|A| < p/9 + o(p)$\mathrm{.}$(iii)$If$|A| \gg \frac{\log\log{p}}{\sqrt{\log{p}}}p$,then$|A + A^*| \geqslant (1 - o(1))\min(2\sqrt{|A|p}, p)$. Here the constants $1\mathrm{/}8$, $1\mathrm{/}9$, and $2$ are the best possible. The proof involves \emph{wrappers}, subsets of a finite abelian group $G$, with which we `wrap’ popular values in convolutions $A\ast B$ for dense sets $A, B \subseteq G$. These objects carry some special structural features, making them capable of addressing both additive-combinatorial and enumerative problems.