<s>[INST] Show that $\frac{1}{a}+\frac{1}{b}\not=\frac{1}{a+b}$ 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?
 [/INST] 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
$$
\frac 1 a + \frac 1 b = 1 + 1 =2 \quad\text{and}\quad\frac 1 {a+b} = \frac 1 2\quad\text{and}\quad 2\ne \frac 1 2.
$$
</s>