Symbolic Logic statement How can I say that "Everybody likes at most one person."  I thought it was 
∀x∀y∃z(Lxy -> y=z)
but this is true even if x likes as many people as you like so this can't be right.  
 A: If $Lxy$ means $x$ likes $y$, then the sentence  $$\forall x\forall y\forall z((Lxy\wedge Lxz)\rightarrow y=z)$$ is the logical formulation of the statement. In words, it says that for every $x$, if $x$ likes some $y$ and some $z$, then $y$ and $z$ should be the same. 
A: You write:

I thought it was
$\forall x \forall y \exists z(Lxy \to y=z)$
but this is true even if x likes as many people as you like so this can't be right.

You are correct that that is not right, and indeed this does not rule out multiple $y$'s that can be liked by $x$, because for every $y$ that is liked by $x$, you can simply point to $z=y$ as a $z$ for which it is true that $z=y$!  In other words, the addition of the $\exists z$ does absolutely nothing to restrict the $y$'s liked by $x$
However, what I think you were trying to do is:
$\forall x \exists z \forall y (Lxy \to y=z)$
And that one works! Now you are are saying that there is something $z$ that any $y$ liked by $x$ must equal to. In other words, you cannot have two different $y$'s both being liked by $x$, because both must equal this one $z$, and hence cannot be different. So, this statement would indeed imply that $x$ can like at most one person.
Moreover, this statement does not say that there actually is a person that $x$ likes. You might think so, because of the existential, but notice that if no one is liked by $x$, then the antecedent of the conditional is always false, and the whole conditional true, and hence we can point to any $z$ to make the statement vacuously true. So, as long as the domain is not empty (which is what most logics inherently assume), this statement would be true even if $x$ doesn't like anyone.
In sum, this statement captures exactly what we want: that every $x$ likes at most one person.
You were very close!
