Find the minimum value of $x^2 + 4xy + 4y^2 + 2z^2$ given that $xyz = 32$

Question

I had a go at this question and believed my answer and solution to be correct, but it turned out not to be.

Here it is: ($x, y, z$ are all greater than 0.)

This solution gives the incorrect answer and I not too sure why.

Any help would be much appreciated!

Thanks

Answer 1

Correct me if I'm wrong but it appears you evaluated the AM-GM inequality incorrectly. By AM-GM:

$$\frac{x^2+4xy+4y^2}{3}\ge\sqrt[3]{16x^3y^3}$$ $$x^2+4xy+4y^2\ge6xy\sqrt[3]{2}$$

Answer 2

HINT:

As $x,y,z>0$ using AM,GM inequality,

$$\frac{x^2+2xy+4y^2+2xy+z^2+z^2}6\ge(x^2\cdot 2xy\cdot 4y^2\cdot 2xy\cdot z^2\cdot z^2)^{\dfrac16}$$

Answer 3

The problem is the notion that the minimum value of $(x+2y)^2$ is $8xy.$ That is only true once you know $xy$. The infimum of all values $(x+2y)^2$ in the domain $xyz=32, x,y,z>0$ is clearly zero.

Alternatively, you assumed $(x+2y)^2=8xy$, but that's only true when $x=2y$, which is a restriction you dropped in the remainder of your answer.

Now, given $z$, we have $xy=\frac{32}{z}$, so $(x+2y)^2 \geq 8xy = \frac{256}{z}$. So we are trying to minimize:

$$f(z)=2z^2 +\frac{256}{z}$$

Answer 4

Use AM-GM!

$x^2+4xy+4y^2+2z^2=x^2+2xy+2xy+4y^2+z^2+z^2\geq 6\sqrt[6]{x^2(2xy)^24y^2(z^2)^2}=6\sqrt[6]{16(xyz)^4}=96.\\$Equality if and only if $x=z=4$ and $y=2$.