Is there a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$? I'm asked in an exercise from an algebra textbook if there exists a pair of numbers $a,b\in\Bbb{R}$ such that $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$
I'm trying to prove that such pair of numbers does not exist, but I'm not sure my proof and my reasoning are correct. Could anyone please check my proof attempt?
First of all, $\exists\frac{1}{a+b}\in\Bbb{R}\iff a\neq -b$
and $\exists(\frac{1}{a}+\frac{1}{b})\in\Bbb{R}\iff a\neq 0 \text{ and } b \neq 0$
$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b} \iff \frac{1}{a+b}=\frac{a+b}{ab}$
$\iff \frac{ab}{a+b}=a+b$
$\iff ab=(a+b)^2$
$\iff ab= a^2 + 2ab + b^2$
$\iff 0 = a^2+2ab+b^2-ab$
$\iff 0= a^2+b^2+ab$
Since $a \neq 0$ and $b \neq 0$; $\exists (ab)^{-1}\in\Bbb{R}$. This allows me to continue like this:
$ 0= a^2+b^2+ab \iff  0(ab)^{-1}=(a^2+b^2+ab)(ab)^{-1}$
$\iff 0 = \frac{a^2+b^2}{ab} + 1$

$\iff -1 = \frac{a^2+b^2}{ab}$    

$a^2+b^2 > 0$ because $a \neq 0$ and $b \neq 0$
If a and b have the same sign, then $ab>0$.
So $\frac{a^2+b^2}{ab}$ could be negative only if: 


*

*$a>0 \text{ and } b<0$ or

*$a<0 \text{ and } b>0$


Taking into account that $a \neq -b$, any of these two options would be possible only if either:


*

*$|a| > |b|$

*$|b| > |a|$


Assuming that $|a| > |b|$ we have the following:
$|a| > |b| \implies |a|*|a| > |b|*|a|$
$\implies |a|^2 > |ab|$
$\implies \frac{|a|^2}{|ab|}>1$
$\implies |\frac{a^2}{ab}|>1$
And $b \neq 0$ so we also have $|\frac{b^2}{ab}|>0$. Therefore $|\frac{a^2}{ab}|+|\frac{b^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions. 

This implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$.

Assuming $|b|>|a|$ we arrive at the same conclusion:
$|b| > |a| \implies |b|*|b| > |a|*|b|$
$\implies |b|^2 > |ab|$
$\implies \frac{|b|^2}{|ab|}>1$
$\implies |\frac{b^2}{ab}|>1$
And $a \neq 0$ so we also have $|\frac{a^2}{ab}|>0$. Therefore $|\frac{b^2}{ab}|+|\frac{a^2}{ab}|> 1 + 0$.
$\frac{a^2}{ab}$ and $\frac{b^2}{ab}$ have the same sign, because the denominator ab determines the sign of both fractions. 

So this also implies $|\frac{a^2}{ab}|+|\frac{b^2}{ab}| = |\frac{a^2}{ab}+\frac{b^2}{ab}| > 1 \implies \frac{a^2+b^2}{ab} > 1 \text{ or } \frac{a^2+b^2}{ab} < -1 \implies \frac{a^2+b^2}{ab} \neq -1$. 

$\blacksquare$
Is this correct??


$\iff 0= a^2+b^2+ab$

Multiplying by $a-b$ gives $a^3-b^3=0 \iff a^3 = b^3\,$, then taking the (real) cube roots $a=b\,$. But it is readily verified that $\frac{1}{a+a} \ne \frac{1}{a} + \frac{1}{a}\,$ for any $\forall a \in \mathbb{R}\,$, so there are no real solutions.
