Let $a, b, c$ be real numbers. Prove that three roots of the equation
\begin{align*}
\frac{b+c}{x-a} + \frac{c+a}{x-b} + \frac{a+b}{x-c} = 3
\end{align*}
are real.
Solution: The solution which the book lists is actually much more elegant than what I will present here. We let $s = a+b+c$ to be the sum. Inserting this into the equation, we see that
\begin{align*}
\frac{s-a}{x-a} + \frac{s-b}{x-b} + \frac{s-c}{x-c} = 3
\end{align*}
Now, multiplying both sides by $(x-a)(x-b)(x-c)$ and rearranging results in
\begin{align*}
(s-a)(x-b)(x-c) + (s-b)(x-a)(x-c) + \\
(s-c)(x-a)(x-b) – 3(x-a)(x-b)(x-c) = 0.
\end{align*}
By inspection, it’s pretty clear that $x = s$ is a root, hence if we find one additional real root we are done as imaginary roots must exist in pairs.
Here is where I deviated against the solutions. I plugged in $x = a, b, c$ respectively and considered all the cases separated by magnitude and showed that there must exist a pair to the left/right of $s$ such that it is positive and negative, hence there exists another real root.