Is it possible to have three real numbers that have both their sum and product equal to $1$? I have to solve $ x+y+z=1$  and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers? 
Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$? 

If any are zero the product is zero.  If all are positive all are less than 1 and the product is less than one so at least one is negative.  If three or one are negative then the product is negative so two are negative and one is positive. 
Wolog $x \le y < 0; z = 1-(x + y)$ or lets relable $y = -a; a > 0; x = -a - e; e \ge 0; z = a+ e + 1$
So we must solve $a(a+e)(1+a+e) = 1$ so $a(e^2 + (1+2a)e + a(1+a) - \frac 1a) = 0$ so $e = \frac{-1-2a \pm \sqrt{4a^2+4a + 1 - 4(a(1+a) - \frac 1a})}{2}= \frac{-1-2a + \sqrt{ 1 + \frac 4a}}{2}$ with the condition $a > 0$ and $1+ \frac 4a \ge (1 + 2a)^2$ or $a^3 + a^2 \le 1$ of which there are no doubt an infinite number of solutions.
