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Let $x,$ $y,$ and $z$ be positive real numbers such that $xyz = 1.$ Find the minimum value of $$(x + 2y)(y + 2z)(xz + 1).$$


I am pretty sure this problem either uses AM-GM or Rearrangement Inequality, but I don't know how to solve it. Can anyone give me a hint?

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  • $\begingroup$ I solved your problem by another way. If you want to see my solution, show please your attempts. $\endgroup$ May 15 '20 at 6:13
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ALERT: this is my solution, not a hint. $$\underbrace{(x+2y)}_{2AM \ge 2GM} \ \ \underbrace{(y+2z)}_{2AM \ge 2GM} \ \ \underbrace{(xz+1)}_{2AM \ge 2GM}\ge 2\sqrt {x \cdot2y} \cdot 2\sqrt {y \cdot2z} \cdot 2 \sqrt{xz} = 8 \sqrt{4x^2y^2z^2}=16xyz=16 $$

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Using the AM-GM inequality we have that :

$(x+2y)\geq 2 \sqrt{2xy}$

$(y+2z)\geq 2 \sqrt{2yz}$

$(xz+1)\geq 2 \sqrt{xz}$

Then $(x+2y) (y+2z) (xz+1) \geq 8 \sqrt{4 (xyz)^2}\implies (x+2y)(y+2z)(xz+1) \geq 16$

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  • $\begingroup$ Well, you basically repeated what I just said :) $\endgroup$ May 15 '20 at 5:15
  • $\begingroup$ I think we answered almost simultaneously, I didn't see that it already had an answer when I posted mine, when I started writing the solution there wasn't an answer $\endgroup$
    – Physmath
    May 15 '20 at 5:18
  • $\begingroup$ Oh, alright :), sorry $\endgroup$ May 15 '20 at 5:19

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