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**rimi****Member**- Registered: 2006-02-21
- Posts: 17

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.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,603

To do superscripts:

`u[sup]n[/sup]`

and you get:

u[sup]n[/sup]

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

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,092

Hence, the question would be.....

Prove that u[sup]n[/sup] x v[sup]1-n[/sup]≤nu+(1-n)v

where

n∈(o,1) and u,v > 0

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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

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

So u[sup]n[/sup] x v[sup]1-n[/sup] ≤ 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.

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**rimi****Member**- Registered: 2006-02-21
- Posts: 17

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

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**rimi****Member**- Registered: 2006-02-21
- Posts: 17

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Rimiq please, help us more.

IPBLE: Increasing Performance By Lowering Expectations.

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