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

1. ParthKohli

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

2. ganeshie8

what parametric form.. polar ?

3. ParthKohli

Nah. Never mind.

4. ganeshie8

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

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

6. anonymous

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

7. ganeshie8

Brilliant!

8. jtvatsim

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

9. ganeshie8

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

10. ganeshie8

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11. jtvatsim

Oh wow! It's the same basic idea, except I did it for the exponential... LOL :)

12. jtvatsim

See here.

13. ganeshie8

14. ganeshie8

That looks so neat! really a very neat substitution x/y = r !

15. jtvatsim

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

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

precisely! Which is exactly how I found it. :)

18. ganeshie8

I see, thats all scratch work which goes in our heads... which we don't normally show in a rigorous proof..

19. anonymous

This reminds me of gamma function

20. ikram002p

a generating function of gamma representation ive seen something like this