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 $x$, $y$, $z$? 
 A: $x+y+z=1$ defines a plane in $\mathbb{R}^3$. 
That plane contains the point $P = (1,0,0)$ where the product equals $0$.
That plane also contains the point $Q = (3,-1,-1)$ where the product equals $3$.
The line segment 
$$\overline{PQ} = \{(1-t)P+tQ \,\bigm|\, 0 \le t \le 1\} = \{(1 + 2t,-t,-t) \, \bigm| \, 0 \le t \le 1\}
$$
lies entirely in that plane. So by the intermediate value theorem, that line segment contains a point where the product equals $1$.
We can get more explicit, finding coordinates of the desired point $(1+2t,-t,-t)$ by setting
$$(1+2t)(-t)(-t)=1
$$
and solving the resulting cubic equation
$$2t^3 + t^2 - 1 = 0
$$
for a root in the interval $0 \le t \le 1$ (which exists, of course, by the intermediate value theorem). I used http://www.tiger-algebra.com to get $t \approx 0.657298088$, and so the point is, approximately,
$$(2.314596176,-0.657298088,-0.657298088)
$$
A: Using Grapher, a few more plots (for $r=1$) to complement the other answers:



A: There are infinite solutions. $z=1-x-y$ is a plane's equation and $z=\dfrac{1}{xy}$ is a complicated curve, but their intersection presents infinitely many real points.

At first I would try some trivial values like $x=0$ or $x=1$ or $x=-1$ and check which of them works and which doesn't.
For instance, if $x=0$, then $xyz=0$ doesn't work.
If $x=1$, $x+y+z=1\rightarrow y+z=0\rightarrow z=-y$ means that $xyz=1\rightarrow y^{2}=-1$ is a complex number also doesn't work.
Then $x=-1$ is a good attempt because $y+z=2$ and $yz=-1$ looks promising.
\begin{equation*}
yz=-1
\end{equation*}
\begin{equation*}
z=-\frac{1}{y}
\end{equation*}
then,
\begin{equation*}
y+z=2
\end{equation*}
\begin{equation*}
y-\frac{1}{y}=2
\end{equation*}
\begin{equation*}
y^{2}-1=2y
\end{equation*}
\begin{equation*}
y^{2}-2y-1=0
\end{equation*}
\begin{equation*}
\Delta=4+4=8
\end{equation*}
\begin{equation*}
y=\frac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}
\end{equation*}
\begin{equation*}
\therefore z=1\mp\sqrt{2}
\end{equation*}
Thus, both $\left(-1,1-\sqrt{2},1+\sqrt{2}\right)$ and $\left(-1,1+\sqrt{2},1-\sqrt{2}\right)$ are valid answers.
I could not comment above for the lack of space. Here are some notes about their signals:
As noticed before none of the variables can be zero, because $xyz\neq0$.
Also the three variables cannot be all positive, because if $x>0$, $y>0$, $z>0$, then it would put them under the interval $0<x,y,z<1$, if one of them would be greater than $1$, for example, $x>1$ then the sum $x+y+z$ would be greater than $1$, since $x>1$, $y>0$, $ z>0$. The problem of that interval is $0<y<1$ would mean that $\dfrac{1}{y}>1$ and for the same reason $0<z<1$ would mean $\dfrac{1}{z}>1$. But $xyz=1$ would make $x=\dfrac{1}{y}\cdot\dfrac{1}{z}>1$ and that would be impossible, so the three variables cannot be all positive and they cannot be localized on that interval.
On the other hand, you cannot have just one negative variable or three negative variables because their product would be negative, so you must have exactly two negative and one positive variables.
A: $$x+y+z=1 \quad \text{and} \quad xyz=1$$
\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 $\xi =\dfrac 13 \left(2 + \sqrt[3]{53 + 6 \sqrt{78}} + \sqrt[3]{53 - 6 \sqrt{78}}\right) \approx 2.314596212276752$ 
A: Consider the polynomial $$P(t)=(t-x)(t-y)(t-z)=t^3-t^2+pt-1$$
Where $p=xy+xz+yz$. The discriminant of the general cubic ($ax^3+bx^2+cx+d$) is $\Delta=b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$, and $\Delta\geq 0$ if and only if all 3 roots are real. The discriminant of $P(t)$ is $\Delta=-4p^3+p^2+18p-31$. Solving the inequality $\Delta\geq 0$ you get $$p\leq\frac{1}{12}\left(1-\frac{217}{\sqrt[3]{6371-624\sqrt{78}}}-\sqrt[3]{6371-624\sqrt{78}}\right)\approx -2.6107$$ So whenever that inequality is satisfied the roots of $t^3-t^2+pt-1$ are real and their sum and product is 1.
A: 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:
$$R(x) = {\sqrt{x^4 - 2x^3 + x^2 - 4x} \over 2x}$$
$$\{x, y, z\} = \left\{x, \frac{1 - x}{2} - R(x), \frac{1 - x}{2} + R(x)\right\}$$
A: 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.
A: General case: $r$ being an arbitrary real number.     
The point $(r,0,0)$ belongs to the plane $x+y+z=r$.
The point $(x,y,z)=((r-a),  \dfrac{a}{2}  ,  \dfrac{a}{2}  )$ also belongs to this plane because $(r-a)+  \dfrac{a}{2}+ \dfrac{a}{2}=r$.        


*

*Consider function  $b=f(a)= (r-a)  \dfrac{a}{2}    \dfrac{a}{2}-r  $.
This  is a continuous cubic function with the parameter $r$ so it has values from $+\infty$ to $-\infty$ and it is assured that this function  crosses line $b=0$ so we have at least one real solution $a_1$ for the equation $(r-a)  \dfrac{a}{2}   \dfrac{a}{2}  =xyz=r$   

*Therefore real solution for the set of equations in the question
always exists .
See picture below for some cases of $r$

A: As there are two equations in three unknowns, you a free to set one unknow.
Now solve
$$\begin{cases}x+y=1-z,\\xy=\dfrac1z,\end{cases}$$
which gives
$$x,y=\frac12\left(1-z\pm\sqrt{(1-z)^2-\frac4z}\right).$$
Solutions exist when the discriminant is non-negative,
$$(1-z)^2-\frac4z\ge0.$$
This is certainly true for values of $z$ above $3$ (more precisely the real root of $z^3-2z^2+z-4=0$), or negative.
For example $-\dfrac32+\sqrt2,-\dfrac32-\sqrt2,4$.
There is no solution with $x,y,z>0$.
A: Those numbers are the intersection of the surface $x y z=1$ and the plane $x+y+z=1$. A quick plot of these surfaces shows that these surfaces do intersect and that the solution consist of a curve (so infinitely many solutions) that looks something like a three-way hyperbola.

A: Another approach: as already noted in other answers, in view of Viète's formulas, you are asking whether there exists $a\in\mathbb{R}$ such that $t^3-t^2+at-1=0$ has three real solutions. 
Equivalently, you are asking whether there is an $a$ such that the graphs of $y=t^3-t^2-1$ and $y=at$ intersect in three points. Equivalently, you are asking whether there is a line through the origin that intersects the graph of the cubic $y=t^3-t^2-1$ in at least two points. 
If you plot this curve it's not complicated to find one graphically, for instance use plot t^3-t^2-1 and 10t in [-4,4] in Wolfram Alpha:

A: A negative fact after many positive solutions: you'll never find a solution with 3 positive numbers! In fact in that case AM-GM holds $$\frac{x+y+z}{3}\geq \sqrt[3]{xyz} $$ which would give $$\frac{1}{3} \geq 1$$
A: Suppose $x=-1/2$ then we have $y + z = 3/2$ and $yz=-2$. We have that $y,z$ are solutions to the quadratic
$$a^2 - 3/2 a - 2.$$
Since the discriminant of this polynomial is positive there are $2$ real solutions (for $y$ and $z$). Hence there exists the required $x,y,z$.
Edit: Note that given any $x <0$ there is $y,z$ such that the $x,y,z$ satisfies your equations. The same proof that I used above works:
Given $x <0$, if $y+z=1-x > 0$ and $yz=\frac{1}{x} < 0$, then $y,z$ are solutions to the quadratic
$$a^2 -(1-x)a + \frac{1}{x}.$$
This quadratic has discriminant $(1-x)^2 -\frac{4}{x} >0.$ Hence there are solutions $x,y,z$ as required. 
Second Edit:
In the case we have $x + y + z = R$ and $xyz=S$ for $R,S \in \mathbb{R}$, $S\ \neq0$ (i.e a generalisation of your edit), we still have solutions. Using the same method as before, consider $y+z=R-x$ and $yz=S/x$. Then, $y,z$ are the roots to the polynomial
$$a^2 -(R-x)a + \frac{S}{x}.$$
As long as we fix $x$ so that $\frac{S}{x} <0$, then the discriminant of this quadratic is positive and we can guarantee a solution set $x,y,z$.
In the case $S=0$, one of $x,y,z$ must be $0$ - solutions are easy to see from there. 
A: Consider cubic
$$ (x-a)(x-b)(x-c) = x^3 - x^2( a+b+c) +x(a b+ b c+ca) -abc $$
Let $ d = (a b + b c+ca) $
$$ x^3 - x^2 + d\,x -  1 =0 $$ 
The roots satisfy your condition. But constant $d$ must be chosen so that all roots are real  by considering its  discriminant in 
Cubic Equation Wiki
resulting  simply as condition:  $ d < -1.2275 $
EDIT1:
(Earlier error in discriminant  $/ d $ )
$$ 4 d^3 - d^2 +18 d -31 <0 ; \quad d < -1.2275 $$
A: 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 $1$.
