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## #1 2006-02-23 16:56:28

rimi
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### Try this one

prove that ,
u (to the power)n × v(to the power)(1-n) ≤nu + (1-n)v
n∈(o,1)  and u,v > 0
by the way how do I bring this superscripts ? As word files are not working.

## #2 2006-02-23 19:07:09

MathsIsFun

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### Re: Try this one

To do superscripts:

#### Code:

`u[sup]n[/sup]`

and you get:
un

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #3 2006-02-23 19:32:17

ganesh
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### Re: Try this one

Hence, the question would be.....
Prove that un x v1-n≤nu+(1-n)v
where
n∈(o,1)  and u,v > 0

Character is who you are when no one is looking.

## #4 2006-02-24 03:51:50

mathsyperson
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### Re: Try this one

If n = 0, then the inequality simplifies down to v ≤ v.

Similarly, if n = 1, then the inequality simplifies down to u ≤ u.

So un x v1-n ≤ nu+(1-n)v all the time because they are actually always equal to each other.

Why did the vector cross the road?
It wanted to be normal.

## #5 2006-02-25 16:28:07

rimi
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### Re: Try this one

No mathsy thats not quite a proof...but I can give you hints if you want...this is simple...indeed

## #6 2006-03-01 16:35:07

rimi
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### Re: Try this one

I guess I can give you some hints about the solution...you can consider the fact that log is a concave function.

## #7 2006-03-01 18:00:37

krassi_holmz
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### Re: Try this one

We must prove that
log(nu+(1-n)v)>=nlog(u)+(1-n)log(v)
Is there some unequation for
log(a+b) and log(a)+log(b)?

IPBLE:  Increasing Performance By Lowering Expectations.

krassi_holmz
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