parmenides51 wrote: Let $x, y, z > 0$ be real numbers such that $xyz + xy + yz + zx = 4$. Prove that $x + y + z \ge 3$. https://artofproblemsolving.com/community/c6h1720820p11130476 Let $x, y, z > 0$ be real numbers such that $xyz + xy + yz + zx = 4$. Prove that $$x+y+z\ge x+\frac{4}{x+1}\ge 3$$

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sqing wrote: parmenides51 wrote: Let $x, y, z > 0$ be real numbers such that $xyz + xy + yz + zx = 4$. Prove that $x + y + z \ge 3$. https://artofproblemsolving.com/community/c6h1720820p11130476 Let $x, y, z > 0$ be real numbers such that $xyz + xy + yz + zx = 4$. Prove that $$x+y+z\ge x+\frac{4}{x+1}\ge 3$$ Probably a very stupid question, but how do I get $x+y+z\geq 3$ from $(\frac{x+y+z}{3})^3 + \frac{(x+y+z)^2}{3} \geq 4$ ?

Either you factor polynomial $f(t)$ where $t=x+y+z$ or estimate you left side using $x+y+z<3$ to reach contradiction.

Let $x+y+z=S$ We can say that $(x+1)(y+1)(z+1)=xyz+xy+yz+zx+x+y+z+1=5+S$ By AM-GM $S+5 \leq \left (\dfrac{S+3}{3}\right)^3$ $\implies$ $108 \leq S^3+9S^2$ It holds for $S \geq 3$ and it doesn't hold for $S <3$