FP1 proof questionWatch

#1
Given that a and b are unequal postive numbers, prove that:

a^3 + b^3 >a^2b + b^2a

So I've messed around to get:

a^2(a-b) + b^2(b-a)>0

but I'm unsure what to do from here.
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11 years ago
#2
(Original post by studienka)
Given that a and b are unequal postive numbers, prove that:

a^3 + b^3 >a^2b + b^2a

So I've messed around to get:

a^2(a-b) + b^2(b-a)>0

but I'm unsure what to do from here.
Now consider what happens if, say a > b
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#3
Okay - if a>b then the first part must be positive.

But then the second part must be negative.

As a>b then a^2>b^2, but how to I wrap up a conclusion here?

I don't want to say "the first part will be more positive than the 2nd part will be negative"

Is there something more mathematical I can wrtie?
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11 years ago
#4
Hint: a^2(a-b) + b^2(b-a) = (a^2-b^2)(a-b)
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11 years ago
#5
(Original post by studienka)
Okay - if a>b then the first part must be positive.

But then the second part must be negative.

As a>b then a^2>b^2, but how to I wrap up a conclusion here?

I don't want to say "the first part will be more positive than the 2nd part will be negative"

Is there something more mathematical I can wrtie?
Well, that's pretty much it. I don't know how anally retentive the examiners are, but I guess if you want to be totally clear then just factorise it as (a+b)(a-b)^2, which is always positive.
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#6
thnx - just what I needed.
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