Show that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$

Question

Problem

Assume that $a,b\in\mathbb{R}-\{0\}$ and that $a+b\not=0$. Prove that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.

My Proof

Let's assume that $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, then it follows that $$ \begin{equation} (a+b)^2-ab=0 \end{equation} $$

Let $x=a+b$ and $y=ab$. Now $b=x-a$ and so $y=a(x-a)=ax-a^2$. The previous equation can be written as $$ x^2-y=0 $$ Substituting $y=ax-a^2$ in this equation gives $$ x^2-ax+a^2=0 $$ The discriminant of this quadratic equation (in $x$) is $-3a^2<0$ and therefore $x=a+b$ has no real solution. This means $a+b\in\mathbb{C}$ and therefore either of $a$ or $b$ or both are not real but this contradicts our assumption that $a,b$ are real numbers. Therefore by contradiction $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$.

My Question

Is my proof correct? are there any alternative proofs?

Answer 1

You have $a^2+ab+b^2=0$. Treating this as a quadratic equation in which the unknown quantity is $a$ gives us this solution: $$ a=\frac{-b\pm\sqrt{b^2 - 4b^2}}{2} = \frac{-b\pm b\sqrt{-3}}{2} = b\left(\frac{-1\pm i\sqrt 3} 2 \right). $$ So if $b$ is any complex number except $0$ (e.g. let $b=1$) and $a$ is as given above, then $$ \frac 1 a + \frac 1 b = \frac 1 {a+b}. $$ But otherwise this last identity does not hold.

Alternatively, one could just seek counterexamples. For example if $a=1$ and $b=1$ then $$ \frac 1 a + \frac 1 b = 1 + 1 =2 \quad\text{and}\quad\frac 1 {a+b} = \frac 1 2\quad\text{and}\quad 2\ne \frac 1 2. $$

Answer 2

Here is another approach:

You have $(a+b)^2 = ab$, so $ab>0$. Expanding the equation gives $a^2+b^2+ab = 0$. Hence we must have $ab<0$, a contradiction.

Answer 3

Assume w.n.l.g. that $a>0$ and $b \in (-a,a]$. Now, if $b>0$, $$ \frac{1}{a} + \frac{1}{b} > \frac{1}{a} > \frac{1}{a+b}. $$ On the other hand, if $b<0$, $$ \frac{1}{a}+\frac{1}{b} \le 0 < \frac{1}{a+b}. $$