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.)

$Notice that <code>$\x^2 + 4xy + 4y^2$</code>$

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

Any help would be much appreciated!

Thanks

Are $x,y,z$ positive? Because you can't use AM/GM if $x+2y=0$, for example... — Thomas Andrews, Jul 15 '14 at 16:56
It's wrong to say that "the minimum value of $(x+2y)^2$ is $8xy$. It is the minimum value, given you know $xy$, but there is no minimum value, in general, and the infimum of $(x+2y)^2$ in this space is $0$. — Thomas Andrews, Jul 15 '14 at 16:59
Sorry, I forgot to add that x, y, z are all greater than 0 therefore, it can't be 0. — Hummus, Jul 15 '14 at 17:02
Related : http://math.stackexchange.com/questions/867793/find-the-maximum-value-of-xy2z3-given-that-x2-y2-z2-1-using — lab bhattacharjee, Jul 16 '14 at 16:39

lab bhattacharjee · Answer 1 · 2014-07-15T17:08:58.943

7

HINT:

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

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

edited Jul 15 '14 at 17:08

answered Jul 15 '14 at 17:01

lab bhattacharjee

274,582

Thank you, I eventually came to this solution, however I am still don't quite understand why my solution is incorrect? Could you offer some help with that? – Hummus Jul 15 '14 at 17:04
@ThomasAndrews: Shouldn't it be $4y^2$ not $y^2$? This also makes it solvable. – Semiclassical Jul 15 '14 at 17:08
@ThomasAndrews, The typo has been rectified – lab bhattacharjee Jul 15 '14 at 17:10
@Semiclassical, Thanks for your observation – lab bhattacharjee Jul 15 '14 at 17:11

Thomas Andrews · Answer 2 · 2014-07-15T17:27:25.773

4

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:

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

edited Jul 15 '14 at 17:27

answered Jul 15 '14 at 17:04

Thomas Andrews

177,126

Thanks for your help! Could you please explain why the infimum of all values is 0, as x,y,and are greater than 0? – Hummus Jul 15 '14 at 17:06
I didn't say minimum, I said "infimum." Given $\epsilon>0$ you can set $$x=y=\frac{\sqrt{\epsilon}}4$$ and then $(x+2y)^2<\epsilon$. – Thomas Andrews Jul 15 '14 at 17:11
Okay, I understand. Thanks for your help! – Hummus Jul 15 '14 at 17:13

score 0 · Answer 3 · answered Jul 15 '14 at 16:58

0

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

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

answered Jul 15 '14 at 16:58

ant11

2,033
15
38

1

No, you are wrong, it is shorthand for $\frac{x+2y}{2}\geq \sqrt{x\cdot 2y}$ implies $(x+2y)^2\geq 8xy$. – Thomas Andrews Jul 15 '14 at 17:01
Ah, I see my mistake, thanks for pointing that out. – ant11 Jul 15 '14 at 17:02

Marcin Malogrosz · Answer 4 · 2019-08-20T12:41:13.597

0

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 $y=2$.

edited Aug 20 '19 at 12:41

answered Jul 15 '14 at 17:18

Marcin Malogrosz

1,960

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

4 Answers4