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0
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avatar+146 

Let  \(x, y,\) and \(z\) be positive real numbers such that  \(xyz=1\)

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

 

Thanks in advance.
 

 Jul 24, 2020
 #1
avatar
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
avatar+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 \(\mathbf{16}\)

 

laugh

 Jul 24, 2020
 #3
avatar
+1

I don't think that 16 is attainable.

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

Guest Jul 24, 2020
 #4
avatar+30931 
+2

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

Alan  Jul 24, 2020
 #5
avatar+146 
+1

Thanks, all of you, it really helped.

 

And the answer is 16.

Firebolt  Jul 27, 2020

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