Question

Inequalities.

Answer 1

If x, y, z are positive real numbers such that \(xyz = 32\), then find the minimum value of \(x^2 + 4xy + 4y^2 + 2z^2\).

Answer 2

\[= (x+2y)^2+2z^2\ge 2 \sqrt{(x+2y)(\sqrt2 z)}\]

Answer 3

oops.

Answer 4

\[= (x+2y)^2 + 2z^2 \ge 2 (x+2y)(\sqrt 2 z) \]

Answer 5

Please don't use Lagrange multipliers...

Answer 6

oh sadness

Answer 7

loll XD

Answer 8

hahaha

Answer 9

I was like halfway there

Answer 10

so you removed all of it then?

Answer 11

yea

Answer 12

well I mean you should have ended what you started

Answer 13

is it ok to assume x, y,z like
x=2^a
y=2^b
z=2^c
where a,b,c are some integers?

Answer 14

Yes, but you can't assume that a, b, c are integers.

Answer 15

+ 1 more thing-> (a+b+c)=5
y can't we do so?

Answer 16

yes, it's obvious that a+b+c=5 since 2^a 2^b 2^c = 2^5
just that we don't know if integers actually minimise the expression

Answer 17

ok then lets take a,b,c to be real numbers

Answer 18

if that's the case, then I think we can complete the problem without taking the 2^x form. just that we still dunno what inequality to use.

Answer 19

\[\frac{(x+2y)^2 + 2z^2}{2} \ge \frac{2(x+2y)^2(2z)^2}{ (x+2y)^2 + 2z^2 } \]not that it helps, but gotta try out all permutations of the am-gm-hm inequality haha

Answer 20

well a,b,c have to be real
so substituting them in the equation we will have this-
\[2\times 2^a+4\times 2^a \times 2^b +2 \times 2^c\]
using AM-GM
\[2^{2a}+2^{2+a+b}+2^{c+1} \ge 2 \sqrt {2^{2(a+b+c)+(a+b)+5}} \]a+b+c=5\[2^{2a}+2^{2+a+b}+2^{c+1} \ge 2\sqrt{2^{(a+b)+15}}\] so we need to find minimum of a+b
since a and b are any real numbers we can take a+b=15
so \[2^{2a}+2^{2+a+b}+2^{c+1} \ge 2\sqrt{2^{15-15}}\]so minimum value occus at equality
so min of\[2^{2a}+2^{2+a+b}+2^{c+1} = 2\]

Answer 21

wait we can get an even more minimum value by putting a+b=-(infty) 0-0 well this is not violating any of the given data??

Answer 22

Here's the solution:\[x^2+2xy+2xy+4y^2+z^2+z^2\ge 6\sqrt[6]{16x^4y^4z^4}=96\]