# Minimum value of $(x + 2y)(y + 2z)(xz + 1)$ when $xyz=1$

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?

• I solved your problem by another way. If you want to see my solution, show please your attempts. May 15 '20 at 6:13

ALERT: this is my solution, not a hint. $$\underbrace{(x+2y)}_{2AM \ge 2GM} \ \ \underbrace{(y+2z)}_{2AM \ge 2GM} \ \ \underbrace{(xz+1)}_{2AM \ge 2GM}\ge 2\sqrt {x \cdot2y} \cdot 2\sqrt {y \cdot2z} \cdot 2 \sqrt{xz} = 8 \sqrt{4x^2y^2z^2}=16xyz=16$$

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

• Well, you basically repeated what I just said :) May 15 '20 at 5:15
• 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 May 15 '20 at 5:18
• Oh, alright :), sorry May 15 '20 at 5:19