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ParthKohli

  • one year ago

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  1. ParthKohli
    • one year ago
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    http://imgur.com/ZsbzWb3

  2. ParthKohli
    • one year ago
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    UUse the weighted means inequality.

  3. dan815
    • one year ago
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    solve this one first |dw:1437428390898:dw|

  4. dan815
    • one year ago
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    u can math it pure, like an optimization problem with a restriction, lagrnge multipled stuff DelF =k-DelG

  5. dan815
    • one year ago
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    but i think there is a neat more intuitive way of thinkign about it this about this problem instead

  6. dan815
    • one year ago
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    im trying to think about it like perimeter problems with max area but iwth a twist

  7. anonymous
    • one year ago
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    AM-GM gives: $$\frac13=\frac13 \left(\frac{\lambda^2}{d_1^2}+\frac{\mu^2}{d_2^2}+\frac{\omega^2}{d_3^2}\right)\ge \sqrt[3]{\frac{\lambda^2\mu^2\omega^2}{d_1^2 d_2^2 d_3^2}}\\\implies (\lambda\mu\omega)^2\le \frac{(d_1 d_2 d_3)^2}{27}\\\implies\lambda \mu \omega\le\frac{d_1 d_2 d_3}{3\sqrt3}$$

  8. anonymous
    • one year ago
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    AM-GM and its derivatives (incl. weighted AM-GM) are consequence of the fact that \(\exp\) is convex, so by Jensen's inequality consider points \(x_i\) and weights \(w_i\) such that \(\sum w_i=1\): $$\exp\left(\sum_i w_i \log x_i\right)\le\sum_i w_i\exp(\log x_i)\\\prod_i \exp(w_i\log x_i)\le\sum_i w_i x_i\\\prod_i x_i^{w_i}\le\sum_i w_i x_i$$

  9. anonymous
    • one year ago
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    for the AM-GM inequality with \(n\) such points choose \(w_i=\frac1n\): $$\prod_i x_i^{1/n}\le\sum_i\frac{x_i}n\\\left(\prod_i x_i\right)^{1/n}\le\frac1n\sum_i x_i\\\sqrt[n]{\prod_i x_i}\le\frac1n\sum x_i$$

  10. ParthKohli
    • one year ago
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    @oldrin.bataku The last term is not \(\omega^2\); it's just \(\omega\). Hence the weights.

  11. dan815
    • one year ago
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    ya i considered AM -GM stuff too

  12. ParthKohli
    • one year ago
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    Y'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 \(a=3,~ b=2, ~c=5\).

  13. dan815
    • one year ago
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    herse 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

  14. ParthKohli
    • one year ago
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    OK, sorta got it using pure AM-GM (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

  15. anonymous
    • one year ago
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    are you sure it's not a typo?

  16. anonymous
    • one year ago
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    the 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 AM-GM so it's actually unnecessary. you are overanalyzing a typo, it's supposed to be \(\omega^2\)

  17. anonymous
    • one year ago
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    also 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): $$\frac14=\frac14\left(\lambda^2/d_1^2+\mu^2/d_2^2+\omega/(2d_3^2)+\omega/(2d_3^2)\right)\ge \sqrt[4]{\frac{\lambda^2\mu^2\omega^2}{4d_1^2d_2^2d_3^4}}\\\implies \lambda^2\mu^2\omega^2\le\frac{d_1^2 d_2^2 d_3^4}{64}\\\implies |\lambda\mu\omega|\le \frac{d_1 d_2 d_3^2}8$$

  18. anonymous
    • one year ago
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    also note that upper bound only works for \(d_1, d_2, \omega\ge 0\), which is where AM-GM 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

  19. ParthKohli
    • one year ago
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    Of course I know that we can apply pure AM-GM if the question says \(\omega^2\). What would the answer have been had it been \(\omega\)?

  20. anonymous
    • one year ago
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    oops, 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 $$|\lambda\mu|\omega\le\frac{|d_1 d_2| d_3^2}8$$

  21. anonymous
    • one year ago
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    if 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

  22. ParthKohli
    • one year ago
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    Oh, ew, of course. I found that on my own too (almost).

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