<s>[INST] Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit. 
Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.

Here is what I have:
$a+b =b+a$ since $R$ is commutative.
Now,
$$(b+a) \cdot b^{-1} = 1+ab^{-1} \\
a^{-1} \cdot (1+ab^{-1}) = a^{-1} + b^{-1}
$$
Thus, $a^{-1} + b^{-1} =a^{-1} \cdot (a+b) \cdot b^{-1}$
Therefore, $(a^{-1} + b^{-1})^{-1} = b \cdot (a+b)^{-1} \cdot a$
And thus, $a^{-1} + b^{-1}$ is a unit in $R$ as well.
Does my answer sound logical. Or are there errors in it?
 [/INST] Your proof is fine.
One way to discover it is by computing freely:
$$
\frac{1}{\dfrac{1}{a}+\dfrac{1}{b}}
=
\frac{1}{\dfrac{a+b}{ab}}
=
\frac{ab}{a+b}
$$
</s>