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.
 A: Important: we know that none of $a,b,c$ can be zero because they all occur in a denominator.                  
a) One could take the product $$(a+b)\cdot(1/a+1/b)=c \cdot 1/c=1$$ Of course this gives $$(a+b)^2=ab \to a^2+b^2 = -ab $$ and thus $a$ and $b$ must have alternating sign.           

b) Or one can take the quotient $$(1/a+1/b)/(a+b)=1/c/c \to 1/(ab)=1/c^2 \to ab=c^2$$ and thus $a$ and $b$ must have the same sign.      


c) By this we find that a) and b) give contrary conditions, $ \to $ Contradiction, $ \to $ no solution.
A: Here's an algebra-free solution, which might complement the sensible approaches above.
You have worked out that $a$ and $b$ are non-zero. So they are either positive or negative.
If they are both positive, then by the first equation $c$ must be greater than both $a$ and $b$, but by the second $c$ must be less than both $a$ and $b$ (reciprocals are monotonically decreasing; one over a big number is always more than one over a small number). So they're not both positive.
If they could both be negative, then the negation of both equations would hold and so we would find an impossible all-positive solution by "taking minus everything".
If one is positive and one negative, you could move the negative one to the other side of each equation, and so again have an impossible all-positive solution.
So all solutions are impossible.
A: Note that
$$
a^2+b^2+ab=a^2+ab+\frac{b^2}{4}+\frac{3}{4}b^2=(a+b/2)^2+\frac{3b^2}{4}\geq 0.
$$
Thus your equality forces $a=b=0$ but this is impossible because we have presumed $a,b,c\neq 0$ to legitimize expressions like $\frac{1}{a},\frac{1}{b},\frac{1}{c}$.
A: Even one more other way: let $a=M+d, b=M-d$ then the product
$$ (a+b) \cdot (\frac 1a+ \frac 1b) = 2M \cdot {2M \over M^2-d^2} =1 = \frac cc $$
Then
$$ 2M \cdot {2M \over M^2-d^2} =1 \\ \phantom a\\ \to 4M^2  =M^2-d^2 \\ \phantom a\\ \to 3M^2 = -d^2  $$
No real solution for $M$ and $d$ except $M=d=0$ but which is excluded by $a,b,c \notin \mathbb \{0\}$

Just for fun one more. We use the geometric mean of $a$ and $b$, such that $a=g \cdot u$, $b=\frac gu $ Then we consider again the product
$$ (a+b) \cdot (\frac 1a + \frac1b) = c \cdot \frac 1c \\
\to g(u+\frac 1u) \cdot \frac 1g (\frac 1u + u) = 1 $$
going to
$$ \to (u+\frac 1u) ^2=1 \tag 1$$
but which is impossible because the absolute value of one of the summands ($u$ or $\frac 1u$) must be greater than 1.
A: Recall the factorization $(a-b)(a^2+ab+b^2)=a^3-b^3$. Since $a$ can't equal $b$ in our situation, what we have is equivalent to $a^3-b^3=0$, or $(a/b)^3=1$. If you have no nontrivial solutions to $x^3=1$ available (such as in the real numbers $\Bbb R$) this would imply $a/b=1$, but we know $a\ne b$ so this can't be true. Hence there are no solutions. Otherwise letting $b=\omega a$ where $\omega$ is nontrivial third root of unity we have that $(a,\omega a,(1+\omega)a)$ (with $a\ne0$) is a solution to the original system.
A: 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.
A: 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.
A: Intuitively, such implication cannot be true for real numbers different from zero, the most simple physical example could be the parallel or series of two resistances, is there any chance that two resistances give the same overall resistance if put in series or in parallel? Of course no, unless they are 0.
However it can be formally shown:
$$\begin{cases} a + b = c \\ \cfrac{1}{a} + \cfrac{1}{b} = \cfrac{1}{c} \end{cases} \; \Rightarrow \; c = a +b = \cfrac{1}{\frac{1}{a} + \frac{1}{b}}$$
Multiplying and dividing both the numerator and the denominator by $a$ and $b$, since $a,b \neq 0$ :
$$a + b = \frac{ab}{a+b}$$
which gives:
$$(a+b)^2 = ab \; \Rightarrow \; a^2+ 2ab + b^2 = ab \; \Rightarrow \; a^2 + ab + b^2 = 0$$
Computing the delta, in view of solving the 2nd degree equation:
$$\Delta = (ab)^2-4a^2b^2 = -3a^2b^2 <0$$
Then the equation has no real solution.  
