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

$$x+y+z=1 \quad \text{and} \quad xyz=1$$
\begin{align}
   xy(1-x-y) &= 1 \\
   xy - x^2y - xy^2 = 1 \\
   xy^2 + x^2y - xy + 1 &= 0 \\
   xy^2 + (x^2 - x)y + 1 &= 0 \\
   y^2 + (x-1)y + \dfrac 1x &= 0 \qquad\text{{$x\ne 0$ since $xyz=1$.}}\\
   y &= \dfrac 12(1-x) \pm \dfrac 12\sqrt{(x-1)^2-\dfrac 4x}
\end{align}
So we need to find non negative real numbers, $n$, for which we can solve
$$(x-1)^2-\dfrac 4x = n^2$$ 
Clearly then, there exists an $n$ for all negative values of x.
When x is positive, Wolfram alpha tells us that the single real-valued solution to $(x-1)^2-\dfrac 4x = 0$ is $x=\xi$, where
$\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$ 
A value of $n$ will exists for all $x \ge \xi$.

Working it out, the solution set is
$$\{x,y,z\} = 
\left\{x,\;
  \dfrac 12(1-x) + \dfrac 12\sqrt{(x-1)^2-\dfrac 4x},\;
  \dfrac 12(1-x) - \dfrac 12\sqrt{(x-1)^2-\dfrac 4x}
\;\right\}$$
for all $x \in (-\infty, 0) \cup [\xi, \infty)$
where $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$ 
