+0

0
93
5
+146

Let  $$x, y,$$ and $$z$$ be positive real numbers such that  $$xyz=1$$

Find the minimum value of $$(x+2y)(y+2z)(xz+1).$$

Jul 24, 2020

#1
0

i think this is wrong....

or right...

using Lagrange multipliers you can show that the minimum occurs when a=b=c=d=1/4 and is equal to 4/3

but you can also argue this from symmetry since they are all required to be positive.

Jul 24, 2020
#2
+25556
+3

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

$$\mathbf{\huge{AM \geq GM }}$$

$$\begin{array}{|rcll|} \hline \dfrac{x+2y}{2} & \ge & \sqrt{x*2y} \\\\ \dfrac{y+2z}{2} & \ge & \sqrt{y*2z} \\\\ \dfrac{xz+1}{2} & \ge & \sqrt{xz*1} \\\\ \hline \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y}\sqrt{y*2z} \sqrt{xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{x*2y*y*2z*xz*1} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & \sqrt{4x^2y^2z^2} \\\\ \dfrac{(x+2y)(y+2z)(xz+1)}{8} & \ge & 2xyz \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*xyz \quad | \quad xyz=1 \\\\ (x+2y)(y+2z)(xz+1) & \ge & 16*1 \\\\ \mathbf{(x+2y)(y+2z)(xz+1)} & \ge & \mathbf{16} \\ \hline \end{array}$$

The minimum value of $$(x+2y)(y+2z)(xz+1)$$ is $$\mathbf{16}$$

Jul 24, 2020
#3
+1

I don't think that 16 is attainable.

My 'messy' calculations point to a minimum of 18.

Guest Jul 24, 2020
#4
+30931
+2

Try x = 2, y = 1, z = 1/2

Alan  Jul 24, 2020
#5
+146
+1

Thanks, all of you, it really helped.

And the answer is 16.

Firebolt  Jul 27, 2020