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I'm trying to solve the following system of equations over positive reals, but I've got stuck. $$ \left\{ \begin{array}{c} ab + bc + ca = 12 \\ a + b + c + 2 = abc \\ \end{array} \right. $$ What I've tried: I've used Vieta's formulas to convert this system of equations to $x^3 + (a_0 + 2)x^2 + 12x + a_0$ by assuming that $a, b, c$ are roots of this polynomial ($a_0$ is just some parameter) and noticed that $abc \le 8$ due to $AM \ge GM$ for ${ab, bc, ca}$.

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your idea was ok, by AM-GM we get with the first equation $$4=\frac{ab+b+c+ca}{3}\geq\sqrt[3]{(abc)^2}$$ from here we get $$abc\le 8$$ and with the second equation we obtain: $$\frac{a+b+c+2}{4}\geq\sqrt[4]{2abc}$$ and with $$\frac{abc}{4}\geq \sqrt[4]{2abc}$$ and from here we get $$abc\geq 8$$ thus $$abc=8$$

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  • $\begingroup$ Therefore $a=b=c=2$ mod basic computation $\endgroup$ – JJR May 8 '17 at 19:05
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Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, the our conditions give $v^2=4$ and $w^3=3u+2$.

In another hand, $$(a-b)^2(a-c)^2(b-c)^2=27(2u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$ which gives $$(u-2)^2(2u+5)(6u+13)\leq0$$ or $u=2$, which gives $w^3=8$ and $a$, $b$ and $c$ are roots of the equation $$x^3-6x^2+12x-8=0$$ or $$(x-2)^3=0,$$ which gives the answer $a=b=c=2$.

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