Symmetric system

It certainly isn't something implied in the solution, but it's not hard to show. Here's the simplest way I personally know of:

Let $x=\frac{2018}{3}, y=\frac{2018}{3}+c, z=\frac{2018}{3}-c$ . Then $x,y,z$ are real and $x+y+z = 2018$ for every real $c$ . Also, for $c \in \left[0,\frac{2018}{3}\right)$ , all the variables are nonzero, so the second equation is equivalent to $f(c) := \frac{3}{2018} + \frac{3}{2018+3c} + \frac{3}{2018-3c} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ Since $f$ is continuous on $\left[0,\frac{2018}{3}\right)$ , $f(0) = \frac{9}{2018} < 1$ and $\displaystyle\lim_{c\to\frac{2018}{3}^-} f(c) = \infty$ , the intermediate value theorem tells us there's a $c^* \in \left[0,\frac{2018}{3}\right)$ such that $f(c^*) = 1$ .

---

Another, slightly more algebraic way, is to recognize by Vieta's formulas that $x,y,z$ are the roots of $X^3 - 2018X^2 + CX - C = 0$ for some nonzero real $C$ $\text{(}C=0$ is valid for the system of equations, but then the expression we're evaluating is undefined). Then we can find the cubic discriminant $\Delta_3$ , which turns out to be a cubic polynomial in $C$ , and since every cubic polynomial is onto as a function $\mathbb{R}\to\mathbb{R}$ , there will be some nonzero real $C$ such that $\Delta_3 > 0$ .