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ParthKohli
 one year ago
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ParthKohli
 one year ago
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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0UUse the weighted means inequality.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0solve this one first dw:1437428390898:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0u can math it pure, like an optimization problem with a restriction, lagrnge multipled stuff DelF =kDelG

dan815
 one year ago
Best ResponseYou've already chosen the best response.0but i think there is a neat more intuitive way of thinkign about it this about this problem instead

dan815
 one year ago
Best ResponseYou've already chosen the best response.0im trying to think about it like perimeter problems with max area but iwth a twist

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0AMGM gives: $$\frac13=\frac13 \left(\frac{\lambda^2}{d_1^2}+\frac{\mu^2}{d_2^2}+\frac{\omega^2}{d_3^2}\right)\ge \sqrt[3]{\frac{\lambda^2\mu^2\omega^2}{d_1^2 d_2^2 d_3^2}}\\\implies (\lambda\mu\omega)^2\le \frac{(d_1 d_2 d_3)^2}{27}\\\implies\lambda \mu \omega\le\frac{d_1 d_2 d_3}{3\sqrt3}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0AMGM and its derivatives (incl. weighted AMGM) are consequence of the fact that \(\exp\) is convex, so by Jensen's inequality consider points \(x_i\) and weights \(w_i\) such that \(\sum w_i=1\): $$\exp\left(\sum_i w_i \log x_i\right)\le\sum_i w_i\exp(\log x_i)\\\prod_i \exp(w_i\log x_i)\le\sum_i w_i x_i\\\prod_i x_i^{w_i}\le\sum_i w_i x_i$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the AMGM inequality with \(n\) such points choose \(w_i=\frac1n\): $$\prod_i x_i^{1/n}\le\sum_i\frac{x_i}n\\\left(\prod_i x_i\right)^{1/n}\le\frac1n\sum_i x_i\\\sqrt[n]{\prod_i x_i}\le\frac1n\sum x_i$$

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0@oldrin.bataku The last term is not \(\omega^2\); it's just \(\omega\). Hence the weights.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ya i considered AM GM stuff too

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Y'know, sometimes, you can see the proportion of the weights. Like in this question: If \(a+b + c= 10\), what is the maximum value of \(a^3 b^2 c^5\)? It's obvious that \(a=3,~ b=2, ~c=5\).

dan815
 one year ago
Best ResponseYou've already chosen the best response.0herse the one with your domain http://www.wolframalpha.com/input/?i=max+xyz%2C+x^2%2Fpi^2%2By^2%2Fe^2%2Bz%2FGoldenRatio^2+%3D+1

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0OK, sorta got it using pure AMGM (but weighted in disguise):\[\dfrac{\lambda^2/3d_1^2 + \lambda^2/3d_1^2 + \lambda^2/3d_1^2 + \mu^2/3d_2^2 + \mu^2/3d_2^2 + \mu^2/3d_2^2 + \nu/6d_3^2 + \cdots +\nu/6d_3^2}{12}\]\[\ge \sqrt[12]{\frac{\lambda^6\mu^6 \nu^6}{3^6 6^6 d_1^6 d_2^6 d_3^6}}\]\[\Rightarrow (1/12) \cdot \sqrt{18} \cdot \sqrt{d_1d_2d_3 }\ge \sqrt{\lambda \mu \nu}\]\[\Rightarrow \lambda\mu\nu \le \frac{d_1d_2d_3}{8}\]Of course the answer is wrong. :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are you sure it's not a typo?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the point of 'weighted' here is simply because the factors \(1/d_1^2,1/d_2^2,1/d_3^2\) are the weights for \(\lambda^2,\mu^2,\omega^2\), but we can just absorb them into the variables for AMGM so it's actually unnecessary. you are overanalyzing a typo, it's supposed to be \(\omega^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also your work is incorrect since you should actually get \(d_3^{12}\), and it's very much overcomplicated (you don't need to expand each in multiples of 3 terms, only the last in twice as many as the others): $$\frac14=\frac14\left(\lambda^2/d_1^2+\mu^2/d_2^2+\omega/(2d_3^2)+\omega/(2d_3^2)\right)\ge \sqrt[4]{\frac{\lambda^2\mu^2\omega^2}{4d_1^2d_2^2d_3^4}}\\\implies \lambda^2\mu^2\omega^2\le\frac{d_1^2 d_2^2 d_3^4}{64}\\\implies \lambda\mu\omega\le \frac{d_1 d_2 d_3^2}8$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also note that upper bound only works for \(d_1, d_2, \omega\ge 0\), which is where AMGM holds; if you allow negatives, then you can make \(x,y\) as large as possible and then \(z\) sufficiently big to satisfy the inequality, whichc learly suggests no upper bound

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Of course I know that we can apply pure AMGM if the question says \(\omega^2\). What would the answer have been had it been \(\omega\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, i meant \(\lambda^2,\mu^2\) big as possible and then \(\omega\) sufficiently big to maintain the inequality; you would need the condition \(\omega\ge 0\) for there to be a maximum for the reason i gave above, but if you had that, then the maximum would be $$\lambda\mu\omega\le\frac{d_1 d_2 d_3^2}8$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if you have everything nonnegative then it gives $$\lambda \mu\omega\le\frac{d_1 d_2 d_3^2}8$$ by the argument i gave

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Oh, ew, of course. I found that on my own too (almost).
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