Tilting the math operator changes nothing, right?

The expression $\frac{1}{a} + \frac{1}{b} = \frac{1}{a + b}$ can be rearranged to $a^2 + ab + b^2 = 0$ (for $a \neq 0$ , $b \neq 0$ , and $a \neq -b$ ).

Using the quadratic equation to solve for $a$ gives a discriminant of $-3b^2$ , which is negative for all real values of $b$ , which means $a$ has no real solutions for any real value of $b$ .

Suppose there exist real $a, b$ such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{a + b}$ . Clearly, $a,b \ne 0$ .

First, suppose $ab > 0$ . There are two cases:

$\begin{array}{rclrcll} \boldsymbol{a > 0, b > 0:} &\implies& a + b > a &\implies& \dfrac{1}{a + b} < \dfrac{1}{a} < \dfrac{1}{a} + \dfrac{1}{b} & \rightarrow\!\leftarrow \\ \\ \boldsymbol{a < 0, b < 0:} &\implies& a + b < a &\implies& \dfrac{1}{a} + \dfrac{1}{b} < \dfrac{1}{a} < \dfrac{1}{a + b} & \rightarrow\!\leftarrow \end{array}$

Next, suppose $ab < 0$ ; WLOG, let $a < 0, b > 0$ . There are two cases:

$\begin{array}{rcll} \boldsymbol{|a| > |b|:} &\implies& \dfrac{1}{a + b} < 0 < \dfrac{1}{a} + \dfrac{1}{b} & \rightarrow\!\leftarrow \\ \\ \boldsymbol{|a| < |b|:} &\implies& \dfrac{1}{a} + \dfrac{1}{b} < 0 < \dfrac{1}{a + b} & \rightarrow\!\leftarrow \end{array}$

Therefore our original assumption must have been false.

First note that we will require that $a,b \ne 0$ and $a \ne -b$ . Then

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

Method 1: By the AM-GM inequality $a^{2} + b^{2} \ge 2 |ab| \gt |ab| \ge -ab$ , and thus $a^{2} + b^{2} = -ab$ has no real solutions.

Method 2: Divide $a^{2} + b^{2} = -ab$ through by $b^{2}$ to end up with $\left(\dfrac{a}{b}\right)^{2} + \dfrac{a}{b} + 1 = 0$ , which is quadratic in $\dfrac{a}{b}$ with solutions $\dfrac{a}{b} = \dfrac{-1 \pm i\sqrt{3}}{2}$ , which are complex, implying that at least one of $a,b$ must be complex, i.e., $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{a + b}$ has no real solutions $a,b$ .

Note: From the 2nd method we have $\dfrac{a}{b} = \large e^{i \frac{2 \pi}{3}}$ and $\dfrac{a}{b} = \large e^{i \frac{4 \pi}{3}}$ as principal roots, so for any non-zero real $b$ we have $\large a = b e^{i \frac{2\pi}{3}}$ or $\large a = b e^{i \frac{4 \pi}{3}}$ .