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ganeshie8
 one year ago
show that \[\large x^ty^{1t}\le tx+(1t)y\]
\(x,y\gt 0\) and \(0\lt t\lt 1\)
ganeshie8
 one year ago
show that \[\large x^ty^{1t}\le tx+(1t)y\] \(x,y\gt 0\) and \(0\lt t\lt 1\)

This Question is Closed

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1I can't help but notice the similarity between this and the parametric form of a complex number.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1what parametric form.. polar ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I think you're referring to equation of straight line between two complex numbers in parametric form.. then there is some similarity yeah :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Take logarithm ofboth sides, note that the log function is concave and rest is simply the statement of Jensen's inequality.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is, \(t\log x + (1t)\log y \leq \log (tx + (1t)y)\)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.2I think I found an "intuitive" proof using some basic Calculus... will post a pdf soon. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Please.. :) that geometric proof using Jensen's inequality is pretty neat, but im not really sure how popular that inequality is...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439361518130:dw

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.2Oh wow! It's the same basic idea, except I did it for the exponential... LOL :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Interesting... please do share

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1That looks so neat! really a very neat substitution x/y = r !

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.2It 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. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Its not so contrived actually... we have \(x^ty^{1t}\le tx+(1t)y\) since \(y\gt 0\), dividing it through out gives \(x^ty^{t}\le t\frac{x}{y}+(1t)\) which is same as \(\left(\frac{x}{y}\right)^t\le t\left(\frac{x}{y}\right)+(1t)\)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.2precisely! Which is exactly how I found it. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1I see, thats all scratch work which goes in our heads... which we don't normally show in a rigorous proof..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This reminds me of gamma function

ikram002p
 one year ago
Best ResponseYou've already chosen the best response.0a generating function of gamma representation ive seen something like this
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