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#1 2006-08-17 01:52:16

nghoihin1
Guest

Final question - Prove

This is the last question! Thx!
Let a and b be positive real numbers such that a^2 + b^2 = 1. Prove that a^4 + b^4 >= 0.5

#2 2006-08-17 06:33:31

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Final question - Prove

Edit: The proof below is invalid, as

But I'm leaving it here in case it sparks any other ideas.

Without loss of generality, we can let a ≥ b.

Note that if a² = .5, then b² = .5.  Also note that if b² > .5, it must be that a² < .5.  But this means that a < b, contradicting a ≥ b.  Thus, we can let a² ≥ .5 ≥ b².

Since b^4 must be non-negative, a^4 + b^4 ≥ .5.  QED


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-08-17 06:46:26

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Final question - Prove

Second try:

With the same reasoning from the previous proof, let:

a² = .5 + ∈, where ∈ ≥ 0.

Then a² + b² = .5 + ∈ + b² = 1.  So b² = .5 - ∈.  So:

And

Thus,

Since ∈² ≥ 0, a^4 + b^4 ≥ .5.  QED

Edit: Interesingly enough, given a^2 (or just a), you can easily calculuate exactly how much greater than .5 a^4 + b^4 will be, without ever having to know what b is or rasing anything to a 4th power.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2006-08-21 07:40:13

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Final question - Prove

Another:

, so
.
Plugging in:

Last edited by krassi_holmz (2006-08-21 07:41:38)


IPBLE:  Increasing Performance By Lowering Expectations.

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