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ParthKohli
 one year ago
Inequalities.
ParthKohli
 one year ago
Inequalities.

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1If x, y, z are positive real numbers such that \(xyz = 32\), then find the minimum value of \(x^2 + 4xy + 4y^2 + 2z^2\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[= (x+2y)^2+2z^2\ge 2 \sqrt{(x+2y)(\sqrt2 z)}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[= (x+2y)^2 + 2z^2 \ge 2 (x+2y)(\sqrt 2 z) \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Please don't use Lagrange multipliers...

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I was like halfway there

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1so you removed all of it then?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1well I mean you should have ended what you started

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.0is it ok to assume x, y,z like x=2^a y=2^b z=2^c where a,b,c are some integers?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Yes, but you can't assume that a, b, c are integers.

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.0+ 1 more thing> (a+b+c)=5 y can't we do so?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.0ok then lets take a,b,c to be real numbers

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1if 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{(x+2y)^2 + 2z^2}{2} \ge \frac{2(x+2y)^2(2z)^2}{ (x+2y)^2 + 2z^2 } \]not that it helps, but gotta try out all permutations of the amgmhm inequality haha

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.0well 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 AMGM \[2^{2a}+2^{2+a+b}+2^{c+1} \ge 2 \sqrt {2^{2(a+b+c)+(a+b)+5}} \]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 \[2^{2a}+2^{2+a+b}+2^{c+1} \ge 2\sqrt{2^{1515}}\]so minimum value occus at equality so min of\[2^{2a}+2^{2+a+b}+2^{c+1} = 2\]

ytrewqmiswi
 one year ago
Best ResponseYou've already chosen the best response.0wait we can get an even more minimum value by putting a+b=(infty) 00 well this is not violating any of the given data??

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Here'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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