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Find all positive real numbers $x,y,z$ which satisfy the following equations simultaneously. $x^3+y^3+z^3=x+y+z$
$x^2+y^2+z^2=xyz$

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    $\begingroup$ Hi, and welcome! Can you please share your thoughts on the problem, and explain what's giving you difficulty? This will help people write responses that are appropriate to your question. $\endgroup$
    – user61527
    Nov 9 '13 at 2:34
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    $\begingroup$ I am unable to find the values of x,y,z $\endgroup$
    – lokesh
    Nov 9 '13 at 2:38
  • $\begingroup$ Can you share what you've tried doing? Can you give some context for the problem, and perhaps mention some techniques you've seen or studied for solving systems like this? $\endgroup$
    – user61527
    Nov 9 '13 at 2:38
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    $\begingroup$ Don't waste your time with those that ask you those questions. Wait, and someone will answer you for sure. $\endgroup$
    – OR.
    Nov 9 '13 at 2:58
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    $\begingroup$ @ABC LOL. Sadly, that's so true. $\endgroup$
    – Calvin Lin
    Nov 9 '13 at 3:10
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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!

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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.

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x=y=z=0 is the only real solution but not sure if that counts as it is not positive.

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