Inequalities.

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If x, y, z are positive real numbers such that \(xyz = 32\), then find the minimum value of \(x^2 + 4xy + 4y^2 + 2z^2\).
\[= (x+2y)^2+2z^2\ge 2 \sqrt{(x+2y)(\sqrt2 z)}\]
oops.

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\[= (x+2y)^2 + 2z^2 \ge 2 (x+2y)(\sqrt 2 z) \]
Please don't use Lagrange multipliers...
oh sadness
loll XD
hahaha
I was like halfway there
so you removed all of it then?
yea
well I mean you should have ended what you started
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?
Yes, but you can't assume that a, b, c are integers.
+ 1 more thing-> (a+b+c)=5 y can't we do so?
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
ok then lets take a,b,c to be real numbers
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.
\[\frac{(x+2y)^2 + 2z^2}{2} \ge \frac{2(x+2y)^2(2z)^2}{ (x+2y)^2 + 2z^2 } \]not that it helps, but gotta try out all permutations of the am-gm-hm inequality haha
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\[2^{2a}+2^{2+a+b}+2^{c+1} = 2\]
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??
Here's the solution:\[x^2+2xy+2xy+4y^2+z^2+z^2\ge 6\sqrt[6]{16x^4y^4z^4}=96\]

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