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.

One more solution can't hurt, right?! We are trying to solve
\begin{align*} a+b=c && \frac{1}{a}+\frac{1}{b}=\frac{1}{c} && && (1) \end{align*}
where $a,b,c$ need to be nonzero numbers. As you pointed out, this is 3 unknowns and 2 equations. However, if we rearrange things a bit, we have the equivalent system
\begin{align*}
\frac{a}{c} + \frac{b}{c} = 1 && \frac{c}{a} + \frac{c}{b} =1.
\end{align*}
So you see, if we define $x=\frac{a}{c}$ and $y=\frac{b}{c}$ (two more nonzero numbers), then we get
\begin{align*}
x+y=1 && \frac{1}{x}+\frac{1}{y}=1. && && (2)
\end{align*}
Since the above system has two equations and two unknowns, it is more straightforward to convert it into a quadratic in one variable and see that Equation 2 has no solutions in the real numbers (although there are two complex solution pairs $(x,y)$).
It follows that there are no real solutions to Equation 1, although $(a,b,c) = (\lambda x, \lambda y, \lambda)$ will be a complex solution to Equation 1 whenever $(x,y)$ is a complex solution to Equation 2 and $\lambda$ is a nonzero complex number.
