<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] 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.
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