<s>[INST] For which values of $a$, $b$ and $c$, if $a + b = c$, then $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$? I have a problem in my homework, which I have tried to solve, but I have just ideas, no real mathematical solutions. The problem is the following:

Suppose we have three real numbers $a$, $b$, and $c$ which satisfy the
  equation:
$$a + b = c$$
Is it then also true that:
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$$
or not? Or is it only true for some particular choice of $a$, $b$, and
  $c$, and which would that be?

My ideas:


*

*I noticed immediately that all $a$, $b$ and $c$ must be different from $0$, because otherwise we would have $\frac{1}{0}$ in the second equation, and that's not defined, as everybody knows.

*I tried to form a system of equations with the equations given in the specification of the problem:
$$\begin{cases} a + b = c \\ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} \end{cases}$$
Since we have 3 variables ($a$, $b$ and $c$), I am not sure if this system of equations can bring me to some solution. 
I have tried to replace $a + b$ in the second equation:
$$b(a + b) + a(a + b) = ab$$
$$ba + b^2 + a^2 + ba = ab$$
We can simplify to:
$$ba + b^2 + a^2 = 0$$
Now, I would not know how to proceed, and sincerely I don't know if my solution (ideas) is correct or not, or how far is it from the real solution.
My guess is that there's no values for $a$, $b$ or $c$ such that the 2 equations are valid.
 [/INST] Assume $\frac 1a + \frac 1b = \frac 1c $, We can conclude that
$$ c = \frac 1{\frac 1a + \frac 1b} = \frac{ab}{a + b}.$$
But then from $a + b = c$ we have
$$ a + b = \frac{ab}{a + b}.$$
Multiply both sides by $a + b$:
$$ a^2 + 2ab + b^2 = ab.$$
Subtract $ab$:
$$ a^2 + ab + b^2 = 0.$$
Since our initial assumption implies that $b \neq 0$, we can divide by $b^2$:
$$ \left(\frac ab\right)^2 + \frac ab + 1 = 0.$$
Suppose $\frac ab = x$; the the above says that $x^2 + x + 1 = 0$.
That quadratic equation has a negative discriminant and
no real roots, contradicting the assumption that any such real numbers
$a$ and $b$ exist.
</s>