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.

Firebolt 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.

Guest Jul 24, 2020

#2**+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** \(\mathbf{16}\)

heureka Jul 24, 2020