Solving Simple Inequality: Tips/Suggestions Needed

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    Inequality
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Discussion Overview

The discussion revolves around solving the inequality \((a+b)^{p} \leq a^{p}+b^{p}\) for \(0 < p < 1\) and \(a, b \geq 0\). Participants are exploring methods to approach this problem without using differentiation.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests manipulating the inequality by factoring out \(a\) and rewriting it as \(\left(1+\frac{b}{a}\right)^{p} \leq 1+\frac{b}{a}^{p}\).
  • Another participant questions the applicability of the binomial theorem due to \(p\) not being an integer.
  • A later reply asserts that the binomial theorem can still be applied, indicating that it works in this context.
  • Some participants express a need to research Newton's generalization as a potential method for addressing the inequality.

Areas of Agreement / Disagreement

There is no clear consensus on the applicability of the binomial theorem, with some participants expressing doubt while others assert its validity. The discussion remains unresolved regarding the best approach to the inequality.

Contextual Notes

Participants have not fully explored the implications of using the binomial theorem for non-integer \(p\), and there is uncertainty about the necessary assumptions for applying different mathematical techniques.

autobot.d
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Is there a way to do this without differentiation?

\left(a+b\right)^{p} \leq a^{p}+b^{p}


0<p<1 and a,b\geq 0

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

\left(1+\frac{b}{a}\right)^{p}\leq 1+\frac{b}{a}^{p}

This seems like the way to go but am stuck. Any suggestions? Thanks.
 
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autobot.d said:
Is there a way to do this without differentiation?

\left(a+b\right)^{p} \leq a^{p}+b^{p}


0<p<1 and a,b\geq 0

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

\left(1+\frac{b}{a}\right)^{p}\leq 1+\frac{b}{a}^{p}

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

Use the binomial theorem?
 
p is not an integer though. Not sure binomial thm would work.

0 < p < 1
 
autobot.d said:
p is not an integer though. Not sure binomial thm would work.

0 < p < 1

It works.
 
Did not know that, will do some research.
 
autobot.d said:
Did not know that, will do some research.

It is Newton's generalisation that works. It is there on wikipedia.
 

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