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.
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.
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.