Simple inequality?

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  • #1
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Main Question or Discussion Point

Is there a way to do this without differentiation?

[itex]\left(a+b\right)^{p}[/itex] [itex]\leq a^{p}+b^{p}[/itex]


0<p<1 and a,b[itex]\geq[/itex] 0

pulling the a out of the the first part and dividing by it to get

[itex]\left(1+\frac{b}{a}\right)^{p}[/itex][itex]\leq 1+\frac{b}{a}^{p}[/itex]

This seems like the way to go but am stuck. Any suggestions? Thanks.
 

Answers and Replies

  • #2
371
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Is there a way to do this without differentiation?

[itex]\left(a+b\right)^{p}[/itex] [itex]\leq a^{p}+b^{p}[/itex]


0<p<1 and a,b[itex]\geq[/itex] 0

pulling the a out of the the first part and dividing by it to get

[itex]\left(1+\frac{b}{a}\right)^{p}[/itex][itex]\leq 1+\frac{b}{a}^{p}[/itex]

This seems like the way to go but am stuck. Any suggestions? Thanks.
Use the binomial theorem?
 
  • #3
68
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p is not an integer though. Not sure binomial thm would work.

0 < p < 1
 
  • #4
371
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p is not an integer though. Not sure binomial thm would work.

0 < p < 1
It works.
 
  • #5
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Did not know that, will do some research.
 
  • #6
371
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Did not know that, will do some research.
It is Newton's generalisation that works. It is there on wikipedia.
 

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