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

Suppose there actually is such a triple, $(x,y,z)$, of real numbers.  Let $u$ be a variable and let's force $x$, $y$, and $z$ to be the roots of a polynomial.  We would consider \begin{align*}
    0 &= (u-x)(u-y)(u-z)  \\  
      &= u^3 - (x+y+z)u^2 + (xy + xz + yz) u - (x y z)  \\
      &= u^3 - u^2 + (xy + xz + yz) u - 1  \\
      &= u^3 - u^2 + A u - 1  \text{,}
\end{align*} where we have used the abbreviation $A = xy + xz + yz$.  If we let $A$ range over the complex numbers, we get three (generically complex) roots that satisfy your equation.  Clearly, if $x$, $y$, and $z$ are real, $A$ is real, so we should restrict $A$ to the reals, but this isn't quite strong enough to make $x$, $y$, and $z$ real.  (In fact, if $A$ is greater than about $-2.6$, two of the roots are complex.  The exact bound for $A$ is the one real root of $4A^3 - A^2 - 18 A + 31$, the discriminant of the polynomial in $u$.)
The above says we could take $A = -3$ and then $x$, $y$, and $z$ are \begin{align*}
    -1, 1-\sqrt{2}, \text{ and } 1+\sqrt{2}  \text{.}
\end{align*}  The sum of the two conjugates is $2$ and adding the first, the sum of all three is $1$.  The product of the conjugates is $1 - 2 = -1$, since this is an example of a difference of two squares, and so the product of all three is $1$.
