problem on solving equations with three variables Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously.
    $x^3+y^3+z^3=x+y+z$
    $x^2+y^2+z^2=xyz$
 A: The equations do not have positive solutions. It follows directly from AM-GM inequalities.
Recall that $x,y,z> 0$ then
$x+y+z=x^3+y^3+z^3\geq3xyz=3(x^2+y^2+z^2)$
$3(x^2+y^2+z^2)-(x+y+z)^2=(x-y)^2+(y-z)^2+(z-x)^2\geq 0$
Hence $x+y+z\geq (x+y+z)^2$
then $x+y+z\leq 1$.
But $x,y,z>0$, so $x,y,z$ are strictly smaller than 1.
So $x^3<x$, $y^3<y$, $z^3<z$, which means $x^3+y^3+z^3<x+y+z$, a contradiction!
A: This is an answer that I am writing to give an alternative, brainless, path for these kind of problems, as an answer using means will soon appear (has already appeared).
Check the details as I have to go now.
First we use Newton's identities to write everything in terms of the elementary symmetric polynomials.
Denote 
$e_1=x+y+z$, $p_1=x+y+z$
$e_2=xy+xz+yz$, $p_2=x^2+y^2+z^2$
$e_3=xyz$, $x^3+y^3+z^3$.
Then $e_2=(e_1^2-p_2)/2$
The given equations are $p_3=e_1$ and $p_2=e_3$. So, $e_2=(e_1^2-e_3)/2$
So $Q(t):=(t-x)(t-y)(t-z)=t^3-e_1t^2+\frac{(e_1^2-e_3)}{2}t-e_3$
We also have $p_3=e_1p_2-e_2p_1-3e_3=e_1e_3-\frac{(e_1^2-e_3)}{2}e_1-3e_3$. So, from the given equation, we get $e_1e_3-\frac{(e_1^2-e_3)}{2}e_1-3e_3=e_1$, from where we can solve for $e_3$ to get $e_3=\frac{e_1^3/2}{e_1-e_1/2-3}$.
Then $$Q(t)=t^3-e_1t^2+\frac{(e_1^2-\frac{e_1^3/2}{e_1-e_1/2-3})}{2}t-\frac{e_1^3/2}{e_1-e_1/2-3}$$
Or $$Q(t)=t^3-e_1t^2+\frac{(\frac{-3e_1^2}{e_1/2-3})}{2}t-\frac{e_1^3/2}{e_1/2-3}$$
Now, we try to impose that this polynomial has three positive real roots.
For example: From Descartes' rule of signs, if $e_1/2-3\geq0$ then $Q$ has only one positive root, but it must have 3 or them. So $e_1<3/2$.
We can now compute the discriminant and impose the condition that the roots are reals.
A: x=y=z=0 is the only real solution but not sure if that counts as it is not positive.
