Mistakes give rise to a problem!

$\dfrac{1}{a} + \dfrac{1}{b}=\dfrac{2}{a+b} \\ \Rightarrow\dfrac{a+b}{ab}=\dfrac{2}{a+b} \\ \Rightarrow (a+b)^2=2ab \\ \Rightarrow a^2+2ab+b^2-2ab=0 \\ \Rightarrow a^2+b^2=0$

Since $a^2,b^2\geq 0$ , the only possible solution is $a=b=0$ .But if $a=b=0$ , then the given expression becomes undefined. Hence , the possible $(a,b)$ are $0$ .

This problem can easily be solved using the AM-HM Inequality:

$AM=\frac{a+b}{2}$

$HM=\frac{2ab}{a+b}$

$HM \leq AM \Longrightarrow \frac{2ab}{a+b} \leq \frac{a+b}{2}$

$\frac{1}{\frac{2ab}{a+b}} \geq \frac{1}{\frac{a+b}{2}} \Longrightarrow \frac{a+b}{2ab} \geq \frac{2}{a+b}$

$\frac{1}{2} \Big(\frac{1}{a}+\frac{1}{b}\Big) \geq \frac{2}{a+b}$

We now must break into cases. Let us start by saying that neither $a$ nor $b$ can be $0$ as this would make the fractions on the left side of the equation undefined.

If we restrict $a$ and $b$ to the positive, real numbers we have:

$\frac{1}{a}+\frac{1}{b} > \frac{1}{2} \Big(\frac{1}{a}+\frac{1}{b}\Big) \geq \frac{2}{a+b}$

Meaning there are no solutions when $a$ and $b$ are greater than $0$ . If $a$ and $b$ are both negative we have a similar inequality:

$\frac{1}{a}+\frac{1}{b} < \frac{1}{2} \Big(\frac{1}{a}+\frac{1}{b}\Big) \leq \frac{2}{a+b}$

Our final case to check is if one variable is positive and the other is negative. WLOG we let $a < 0 < b$ . If $|a|<|b|$ , the LHS is negative, while the RHS is positive. Alternatively, if $|a|<|b|$ , the LHS is positive and the RHS is negative.

Finally, we can't have $a=-b$ because the RHS will be undefined, so we're done. These two terms cannot be equal any and all pairs of real numbers, therefore our answer is $\boxed{0}$

multiply each side by $ab(a+b)$ to obtain $a^2+b^2=0$ which obviously have no solution in the range given.

In first place, it is obvious that we should discard any pair that leads to a+b=0. Having it said, we can rewrite the left side of the equation using the properties of proportions (the proportion remains the same if we add the same term to both numerator and denominator) : 1/a = (1+b)/(a+b) and 1/b = (1+a)/(b+a)

Thus, the LHS becomes (a+b+2)/(a+b) = 1 + 2/(a+b)

Substituing in the original expression we get the absurd solution 1=0. Therefore, independently of the values of a and b, the equation has no solution (real one, at least)