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Solve the following system of equations in $\Bbb R^+$: $$ \left\{ \begin{array}{l} xy+yz+xz=12 \\ xyz=2+x+y+z\\ \end{array} \right. $$

I did as follows.
First equation yields to: $(x+y+z)^2-x^2-y^2-z^2=24$. Now, using second equation we get: $x^2y^2z^2-4xyz=20+x^2+y^2+z^2$ . Here I stopped...

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  • $\begingroup$ One solution is $x=y=z=2$, there are others. $\endgroup$ – Orest Bucicovschi Sep 13 '17 at 3:51
  • $\begingroup$ @orangeskid Yes,but how to find the other non Trivial ones? $\endgroup$ – Hamid Reza Ebrahimi Sep 13 '17 at 3:52
  • $\begingroup$ There are infinitely many real solutions. $\endgroup$ – Orest Bucicovschi Sep 13 '17 at 3:54
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    $\begingroup$ Looks like it is indeed the only one. The function $x y + x z + y z$ has a minimum $12$ given the other condition, achieved only at $(2,2,2)$. I think you should prove that. $\endgroup$ – Orest Bucicovschi Sep 13 '17 at 4:03
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    $\begingroup$ Consider first the case of positive numbers. $\endgroup$ – Orest Bucicovschi Sep 13 '17 at 4:06
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The condition gives $$xyz=2+x+y+z$$ or $$xyz=2+\frac{(x+y+z)(xy+xz+yz)}{12}$$ or $$12xyz=24+\sum_{cyc}(x^2y+x^2z+xyz)$$ or $$3xyz=24+\sum_{cyc}(x^2y+x^2z-2xyz),$$ which gives $$3xyz-24=\sum_{cyc}(x^2y+x^2z-2xyz)=\sum_{cyc}z(x-y)^2\geq0.$$ Thus, $xyz\geq8$ and the equality occurs for $x=y=z$.

On the other hand, by AM-GM
$$12=xy+xz+yz\geq3\sqrt[3]{x^2y^2z^2},$$ which gives $xyz\leq8.$

Thus, $xyz=8$ and we get $x=y=z=2$.

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I will show that the only positive solution is indeed $(2,2,2).$ $$\dfrac{xyz}{xy+yz+zx} = \dfrac{1}{6}+\dfrac{x+y+z}{12} = \dfrac{\sqrt{xy+yz+zx}}{12\sqrt{3}}+\dfrac{x+y+z}{12}\geq\dfrac{xyz}{4(xy+yz+zx)}+\dfrac{3xyz}{4(xy+yz+zx)} = \dfrac{xyz}{xy+yz+zx}$$, by AM-GM inequalities provided that all of them are non-negative.

Therefore, either the problem intended to ask for only non-negative solutions, or it has infinitely many solutions indicated by the answer above.

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[ EDIT ] The question has been changed to require $\,x,y,z \ge 0\,$ now, which renders the following not applicable any longer.


(Too long for a comment.) Let $x=a+2, y=b+2, z=c+2\,$, then the conditions rewrite as:

$$\require{cancel} \begin{cases} \begin{align} a b + b c + a c + 4(a + b + c) + \cancel{12} &= \cancel{12} \\ a b c + 2(a b + a c + b c) + 4(a + b + c) + \cancel{8} &= a+b+c + \cancel{8} \end{align} \end{cases} $$

$$\require{cancel} \iff\;\; \begin{cases} \begin{align} a b + b c + a c + 4(a + b + c) &= 0 \\ a b c + 2(a b + a c + b c) + 3(a + b + c) &= 0 \end{align} \end{cases} $$

Let $u = a+b+c\,$, then:

$$\require{cancel} \begin{align} a b + b c + a c &= -4 u \\ a b c &= 5u\end{align} $$

Therefore $a,b,c$ are the roots of $t^3-ut^2-4 ut-5u\,$, and the problem reduces to finding for what values of $u$ the cubic has three real roots (other than $u=0$ which corresponds to $x=y=z=2\,$).

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  • $\begingroup$ I think this way it's getting more complicated! $\endgroup$ – Hamid Reza Ebrahimi Sep 13 '17 at 4:01
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    $\begingroup$ @HamidRezaEbrahimi I don't know that the answer is simple ;-) The cubic has three distinct real roots for $u=-13$ for example, so your system does in fact have non-trivial solutions, infinitely many in fact. $\endgroup$ – dxiv Sep 13 '17 at 4:05
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I will use the cubic equation with roots $x$, $y$, $z$, inspired by the answer of @dxiv:.

$$X^3 - s X^2 + 12 X - (2+s) =0$$

The discriminant of the equation is $-(2s+13)(2s+15)(s-6)^2$. If $s\ge 0$ then the discriminant is positive only for $s=6$, which gives $a=b=c=2$. Therefore: the system has only one positive solution, but infinitely many negative ones, for each $s \in [-\frac{15}{2}, -\frac{13}{2}]$ we get a solution with $a+b+c = s$. For instance, for $s=-\frac{13}{2}$, we get $a=b=-3$, $c=-\frac{1}{2}$ a solution.

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On one hand, we have $$(xy+yz+zx)^2 \geq 3xyz(x+y+z)$$, then one has $$-8 \leq x+ y +z \leq 6.$$

On other hand, we have $$(x+y+z)^2 \geq 3(xy+yz+zx) = 36,$$

then $$x+y+z \geq 6$$ or $$x+y+z \leq -6.$$

So, you have $x+y+z = 6$ or $-8 \leq x+y+z \leq -6$ (but x, y, z is non-negative!).

Thus, $x+y+z = 6$ and $x=y=z$.

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Using AM-GM: $$12=xy+yz+xz\ge 3\sqrt[3]{(xyz)^2} \Rightarrow xyz\le 8 \ \ \ \ (1)$$ $$xyz=2+x+y+z\le8 \Rightarrow x+y+z\le 6 \ \ \ \ \ (2)$$ The equality occurs when $x=y=z=2$.

Using C-S: $$(x^2+y^2+z^2)(y^2+z^2+x^2)\ge (xy+yz+zx)^2 \Rightarrow x^2+y^2+z^2\ge 12 \ \ \ \ (3)$$ The equality occurs when $x=y=z=2$.

From the first equation: $$(x+y+z)^2=24+x^2+y^2+z^2\ge 36 \Rightarrow x+y+z\ge 6 \ \ \ \ \ (4)$$ From $(2)$ and $(4)$, and then from the second equation we get: $$x+y+z=6, \\ xyz=8.$$ So, we get: $$x=y=z=2.$$

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