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 $x$, $y$, $z$? 

Given $x$ and the requirements $x + y + z = xyz = 1$, we can derive possible values for $y$ which in turn fully determines $z$.
Substituting $z = 1 - x - y$ in $xyz = 1$:
$$xy (1 - x - y) = 1$$
$$xy - x^2y - xy^2 = 1$$
$$(-x)y^2 + (x - x^2)y - 1 = 0$$
The quadratic formula now tells us that y is given by:
$$y = {-(x - x^2) \pm \sqrt{(x - x^2)^2 - 4x} \over -2x}$$
$$y = \frac{1-x}{2} \pm {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$
Before tackling the domain, consider this:
$$y = \frac{1-x}{2} \pm R(x)$$
$$z = 1 - x - (\frac{1-x}{2} \pm R(x))$$
$$z = \frac{1 - x}{2} - (\pm R(x))$$
Thus the sign of the square root is irrelevant, as it merely swaps $y$ and $z$ around. Now all that remains is finding the domain for $x$ such that the solution is real. This means $x^4 - 2x^3 + x^2 - 4x > 0$. You can take the time to solve this by hand (it's actually a cubic), or plug it straight into your favourite CAS and find out it's:
$$x < 0 \vee x > \frac{2+\sqrt[3]{53-6\sqrt{78}}+\sqrt[3]{53+6\sqrt{78}}}{3}$$
$$X < 0 \vee x > 2.3146...$$
So the final parametric solution is:
$$R(x) = {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$
$$\{x, y, z\} = \left\{x, \frac{1 - x}{2} - R(x), \frac{1 - x}{2} + R(x)\right\}$$
