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ganeshie8

  • one year ago

show that \[\large x^ty^{1-t}\le tx+(1-t)y\] \(x,y\gt 0\) and \(0\lt t\lt 1\)

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  1. ParthKohli
    • one year ago
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    I can't help but notice the similarity between this and the parametric form of a complex number.

  2. ganeshie8
    • one year ago
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    what parametric form.. polar ?

  3. ParthKohli
    • one year ago
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    Nah. Never mind.

  4. ganeshie8
    • one year ago
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    I think you're referring to equation of straight line between two complex numbers in parametric form.. then there is some similarity yeah :)

  5. anonymous
    • one year ago
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    Take logarithm ofboth sides, note that the log function is concave and rest is simply the statement of Jensen's inequality.

  6. anonymous
    • one year ago
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    That is, \(t\log x + (1-t)\log y \leq \log (tx + (1-t)y)\)

  7. ganeshie8
    • one year ago
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    Brilliant!

  8. jtvatsim
    • one year ago
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    I think I found an "intuitive" proof using some basic Calculus... will post a pdf soon. :)

  9. ganeshie8
    • one year ago
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    Please.. :) that geometric proof using Jensen's inequality is pretty neat, but im not really sure how popular that inequality is...

  10. ganeshie8
    • one year ago
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    |dw:1439361518130:dw|

  11. jtvatsim
    • one year ago
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    Oh wow! It's the same basic idea, except I did it for the exponential... LOL :)

  12. jtvatsim
    • one year ago
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    See here.

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  13. ganeshie8
    • one year ago
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    Interesting... please do share

  14. ganeshie8
    • one year ago
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    That looks so neat! really a very neat substitution x/y = r !

  15. jtvatsim
    • one year ago
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    It looks so contrived, but it arises naturally from playing with the inequality backwards. Assuming the inequality is true and trying to rearrange it leads to this idea. :)

  16. ganeshie8
    • one year ago
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    Its not so contrived actually... we have \(x^ty^{1-t}\le tx+(1-t)y\) since \(y\gt 0\), dividing it through out gives \(x^ty^{-t}\le t\frac{x}{y}+(1-t)\) which is same as \(\left(\frac{x}{y}\right)^t\le t\left(\frac{x}{y}\right)+(1-t)\)

  17. jtvatsim
    • one year ago
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    precisely! Which is exactly how I found it. :)

  18. ganeshie8
    • one year ago
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    I see, thats all scratch work which goes in our heads... which we don't normally show in a rigorous proof..

  19. anonymous
    • one year ago
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    This reminds me of gamma function

  20. ikram002p
    • one year ago
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    a generating function of gamma representation ive seen something like this

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