Cardinality of finite sets: if $B=\{a-b \mid a,b \in A\}$ and $C=\{a+b \mid a,b \in A\}$, then $|C|^2\ge |A||B|$

Question

Let A be a finite set of real numbers. Suppose $B=\{a-b \mid a,b \in A\}$ and $C=\{a+b \mid a,b \in A\}$. Prove that $|C|^2\ge |A||B|$.

I tried to solve this in this way:

We claim that the function $f:C \times C \to A\times B $ such that $f(a+b, c+d) = (a, a-c)$ is surjective, this is because $a$, $b$, $c$, $d$ are all in $A$, so $f$ is surjective, and $|C|^2\ge |A||B|$.

Is that all correct in my demonstration?

Answer 1

No. Your $f$ is not a function.

Suppose $A=\{-1,0,1\}$.

What is $f(0,0)$?

Is it $f(0+0,1+(-1))=(0,-1)$ or $f(1+(-1),0+0)=(1,0)$ or something else?