Finding all the sets of three real numbers that satisfy specific equations I am having trouble finding a set of three numbers of real numbers $(x, y, z)$ satisfying $x + y + z = xy + xz + yz = 3$. I have tried factoring the equations around but I'm not having any luck and I don't think I'm going on the right path.
 A: Given
$$a+b+c=3\tag1$$
$$ab+bc+ca=3\tag2$$
Hint:
$$(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)\tag3$$

Spoiler
From $(3)$ it can easily be shown that
$$a^2+b^2+c^2 = 3$$
Now
$$(a+b+c)^2 = 3^2$$
$$(a+b+c)^2 + a^2+b^2+c^2 - 4(ab+bc+ca) = 3^2 + 3 - 4*3$$
$$a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2 = 0$$
$$(a-b)^2+(b-c)^2+(c-a)^2=0$$
As $a,b,c$ are real numbers and squares of real numbers cannot be negative so
$a-b=b-c=c-a=0;a=b=c$
Replacing in $(3)$ we get
$$a+a+a=b+b+b=c+c+c=3$$
$$a=b=c=1$$
A: Your numbers $a, b, c$ are the roots of
$$
x^3 - 3 x^2 + 3 x - p,
$$
where we take $p = abc$ as a parameter. So
$$
(x -1)^3 = p - 1
$$
and
$$
x = 1 + \omega^i \sqrt[3]{p-1}
$$
are the solutions, where $\omega$ is a primitive third root of $1$.
So if you want $a, b, c$ to be real, I guess you need $p = 1$, so that $a = b = c = 1$.
A: You have two equations in three unknowns, so you should expect to be able to be able to pick one variable as a parameter and solve for the other two.  Since the variables are equivalent, we will take $a$ as the parameter and solve for $b,c$.  We have $$a+b+c=3 \\ ab + bc + ac=3 \\b=3-a-c\\(a+c)(3-a-c)+ac=3\\3a+3c-a^2-2ac-c^2+ac=3\\c^2+(a-3)c+3-3a+a^2=0 \\c=\frac 12\left((3-a)\pm \sqrt{(a-3)^2-4(3-3a+a^2)}\right)\\c=\frac 12 \left((3-a)\pm \sqrt{a^2-6a+9-12+12a-4a^2}\right)\\c=\frac 12 \left((3-a)\pm \sqrt{6a-3-3a^2}\right)\\c=\frac 12 \left((3-a)\pm \sqrt{-3(a-1)^2}\right)$$  The answers are complex unless $a=1$, which shows the only real solution is $a=b=c=1$
A: Try the roots of the polynomials $$x^3-3x^2+3x+r, \  \ r\in\mathbb R.$$
A: Hint:
$a+b+c=3$
$a^2+b^2+c^2=3$
EDITED
From famous Cauchy–Schwarz inequality(http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)
You get: 
$(a^2+b^2+c^2)(c^2+a^2+b^2) \ge(ab+bc+ca)^2$
But the equality holds here,
$(a^2+b^2+c^2)^2-(ab+bc+ca)^2=0$
When we have Cauchy Schwarz inequality: 
$(a_1^2+a_2^2+..+a_n^2)(b_1^2+b_2^2+..+b_n^2) \ge (a_1b_1+a_2b_2+..a_nb_n)^2$
Equality holds iff: 
$(a_1b_2-a_2b_1)=...a_{n-1}b_{n}-a_nb_{n-1}=0$
Apply it over here you get $a^2-bc=b^2-ac=c^2-bc0 \implies a=b=c$
A: Clearly $a=b=c=1$ is a solution.
Let $a=(1-\alpha)$, and $b=(1-\beta)$. Then $a+b+c=3$ implies 
$c = (1 + \alpha + \beta)$
\begin{align}
   ab + bc + ca &= 3 \\
   (1-\alpha)(1-\beta) + 
   (1-\beta)(1 + \alpha + \beta) + 
   (1 + \alpha + \beta)(1-\alpha) &= 3 \\
   -\alpha^2 - \alpha \beta - \beta^2 + 3 &= 3 \\
   \alpha^2 + \alpha \beta + \beta^2 &= 0 \\
   2\alpha^2 + 2\alpha \beta + 2\beta^2 &= 0 \\
   \alpha^2 + \beta^2 + (\alpha^2+2\alpha \beta + \beta^2) &= 0 \\
   \alpha^2 + \beta^2 + (\alpha + \beta)^2 &= 0
\end{align}
Which implies $\alpha = \beta = 0$. So $a=b=c=1$ is the only solution.
