show that $$\large x^ty^{1-t}\le tx+(1-t)y$$

$x,y\gt 0$ and $0\lt t\lt 1$

Question

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

parthkohli · Answer

I can't help but notice the similarity between this and the parametric form of a complex number.

ganeshie8 · Answer

what parametric form.. polar ?

parthkohli · Answer

Nah. Never mind.

ganeshie8 · Answer

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

anonymous · Answer

Take logarithm ofboth sides, note that the log function is concave and rest is simply the statement of Jensen's inequality.

anonymous · Answer

That is, $t\log x + (1-t)\log y \leq \log (tx + (1-t)y)$

ganeshie8 · Answer

Brilliant!

jtvatsim · Answer

I think I found an "intuitive" proof using some basic Calculus... will post a pdf soon. :)

ganeshie8 · Answer

Please.. :) that geometric proof using Jensen's inequality is pretty neat, but im not really sure how popular that inequality is...

ganeshie8 · Answer

|dw:1439361518130:dw|