Solving Simple Inequality: Tips/Suggestions Needed

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The discussion focuses on solving the inequality (a+b)^{p} ≤ a^{p}+b^{p} for 0 < p < 1 and a, b ≥ 0 without using differentiation. A participant suggests manipulating the expression by factoring out 'a' and dividing, leading to the form (1 + b/a)^{p} ≤ 1 + (b/a)^{p}. There is uncertainty about the applicability of the binomial theorem since p is not an integer. However, it is noted that Newton's generalization of the binomial theorem can be utilized in this case. Participants express intent to research this approach further for a solution.
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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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