If $p$ is an odd prime (EDIT: other than 5) for which $-1$ is a quadratic residue mod $p$, and $A$ is the set of non-zero quadratic non-residues mod $p$, then $A+A$ is all of ${\bf Z}/p{\bf Z}$, whilst $A+AA$ is ${\bf Z}/p{\bf Z} \backslash \{0\}$. So counterexamples exist in finite fields, which rules out some methods of proof (e.g. "Ruzsa calculus" by itself will be insufficient). Unfortunately, this example does not appear to be adaptable to the reals (for which $-1$ is certainly not a square, and for which there are no large multiplicative subgroups). Actually it looks difficult to build an example in the complex numbers (or any other characteristic zero field); I don't even see a way to construct an (EDIT: arbitrarily large) finite set $A$ obeying the weaker inequality $|A+AA| < \frac{|A| (|A|+1)}{2}$. One may indeed conjecture (in the spirit of the Erdos-Szemeredi sum-product conjecture) that one always has $|A+AA| \geq \frac{|A| (|A|+1)}{2}$ (EDIT: for sufficiently large $A$), but this is well beyond our current technology to prove. (EDIT: as noted in comments, there are small counterexamples to the weaker inequality, although they do not give counterexamples to the original inequality.)

If $p$ is an odd prime (EDIT: other than 5) for which $-1$ is a quadratic residue mod $p$, and $A$ is the set of non-zero quadratic non-residues mod $p$, then $A+A$ is all of ${\bf Z}/p{\bf Z}$, whilst $A+AA$ is ${\bf Z}/p{\bf Z} \backslash \{0\}$. So counterexamples exist in finite fields, which rules out some methods of proof (e.g. "Ruzsa calculus" by itself will be insufficient). Unfortunately, this example does not appear to be adaptable to the reals (for which $-1$ is certainly not a square, and for which there are no large multiplicative subgroups). Actually it looks difficult to build an example in the complex numbers (or any other characteristic zero field); I don't even see a way to construct an (EDIT: arbitrarily large) finite set $A$ obeying the weaker inequality $|A+AA| < \frac{|A| (|A|+1)}{2}$. One may indeed conjecture (in the spirit of the Erdos-Szemeredi sum-product conjecture) that one always has $|A+AA| \geq \frac{|A| (|A|+1)}{2}$ (EDIT: for sufficiently large $A$), but this is well beyond our current technology to prove. (EDIT: as noted in comments, there are small counterexamples obeying the weaker inequality, although they do not give counterexamples to the original inequality.)

If $p$ is an odd prime for which $-1$ is a quadratic residue mod $p$, and $A$ is the set of non-zero quadratic non-residues mod $p$, then $A+A$ is all of ${\bf Z}/p{\bf Z}$, whilst $A+AA$ is ${\bf Z}/p{\bf Z} \backslash \{0\}$. So counterexamples exist in finite fields, which rules out some methods of proof (e.g. "Ruzsa calculus" by itself will be insufficient). Unfortunately, this example does not appear to be adaptable to the reals (for which $-1$ is certainly not a square, and for which there are no large multiplicative subgroups). Actually it looks difficult to build an example in the complex numbers (or any other characteristic zero field); I don't even see a way to construct a finite set $A$ obeying the weaker inequality $|A+AA| < \frac{|A| (|A|+1)}{2}$. One may indeed conjecture (in the spirit of the Erdos-Szemeredi sum-product conjecture) that one always has $|A+AA| \geq \frac{|A| (|A|+1)}{2}$, but this is well beyond our current technology to prove.

If $p$ is an odd prime (EDIT: other than 5) for which $-1$ is a quadratic residue mod $p$, and $A$ is the set of non-zero quadratic non-residues mod $p$, then $A+A$ is all of ${\bf Z}/p{\bf Z}$, whilst $A+AA$ is ${\bf Z}/p{\bf Z} \backslash \{0\}$. So counterexamples exist in finite fields, which rules out some methods of proof (e.g. "Ruzsa calculus" by itself will be insufficient). Unfortunately, this example does not appear to be adaptable to the reals (for which $-1$ is certainly not a square, and for which there are no large multiplicative subgroups). Actually it looks difficult to build an example in the complex numbers (or any other characteristic zero field); I don't even see a way to construct an (EDIT: arbitrarily large) finite set $A$ obeying the weaker inequality $|A+AA| < \frac{|A| (|A|+1)}{2}$. One may indeed conjecture (in the spirit of the Erdos-Szemeredi sum-product conjecture) that one always has $|A+AA| \geq \frac{|A| (|A|+1)}{2}$ (EDIT: for sufficiently large $A$), but this is well beyond our current technology to prove. (EDIT: as noted in comments, there are small counterexamples to the weaker inequality, although they do not give counterexamples to the original inequality.)