Is the minimum number of relations in a free product, the sum of the minimum number of relations in the free factors? Say $\rho(G)$ is the minimum number of relations required to present the group $G$. Is $\rho(A*B)= \rho(A)+\rho(B)$? What can be said about $\rho(A*B)$? 
A while ago I was thinking about $C_3*C_4$, thinking it was obvious that this group should not be a one-relator group. I am pretty sure at one point I came up with an argument but it seemed awfully complicated, and using specific information about one-relator groups for something that "feels" obvious. 

Using the comments below as a guide, one can prove that if the abelianization of $A,B$ is finite and $\rho(A)=r(A),\rho(B)=r(B)$ where $r(G)$ is the rank of the group (minimum number of generators), then we can get that $\rho(A*B)=\rho(A)+\rho(B)$.


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*It is known that $r(A*B) =r(A)+r(B)$, this can be found in Ch IV Cor 1.9 in Lyndon and Schupp.

*We also have that $\rho(A*B) \leq \rho(A)+\rho(B)=r(A)+r(B)$.

*If $\rho(A*B) < r(A*B)$, then a presentation witnessing that inequality can be shown to have an infinite abelianization, since the number of generators would be strictly greater than the number of relations (see this answer). This contradicts that the abelianization of $A*B$ is finite.


So finite free products of finite cyclic groups works and other groups too.

I could see a careful analysis the abelianization of f.g. groups giving more more information. And it does seem that a somewhat related concept, group deficiency, is related to some cohomology groups (I don't know any cohomology, but I would welcome an answer, even if it used cohomology). 
 A: $\DeclareMathOperator{\def}{def}$
It turns out that there are groups $A*B$ with $\rho(A*B) < \rho(A)+\rho(B)$! 


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*$(C_2 \times C_2) * (C_3 \times C_3)$ has a 5 relation presentation $$\langle x,y,z,t \mid x^2,z^3,xyx^{-1}y^{-3}, ztz^{-1}t^{-4}, y^2t^{-3} \rangle$$ by theorem 3 in Presentation classes, 3-manifolds and free products by Cynthia Hog, Martin Lustig and Wolfgang Metzler.  

*The Schur multiplier  of a finite group $G$ gives information about the deficiency of the group, where the deficiency is the maximum value of generators minus relators over all presentation of the group which will be called $\def G$. In particular, the Schur multiplier can be generated by $-\def G$ generators. Both $C_2 \times C_2$ and $C_3 \times C_3$ have nontrivial Schur multiplier, so neither can have deficiency equal to $0$. That implies that they do not have two relator, two generator presentations. 

*The groups $C_i \times C_i$ is not cyclic and can not have more generators than relations, since they are finite.  So the fewest number of relations comes from the standard presentation $$C_i \times C_i\cong\langle x,y \mid x^i, y^i, xyx^{-1}y^{-1} \rangle.$$


This means that $$\rho {\big{(}}(C_2 \times C_2) *(C_3 \times C_3) {\big )}\leq 5 <6=\rho(C_2 \times C_2) + \rho(C_3 \times C_3).$$
