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If $p,q$ are positive quantities and $0 \leq m\leq 1$ then Prove that $$(p+q)^m \leq p^m+q^m$$

Trial: For $m=0$, $(p+q)^0=1 < 2= p^0+q^0$

and for $m=1$, $(p+q)^1=p+q =p^1+q^1$.

So, For $m=0,1$ the inequality is true.How I show that the inequality is also true for $0 < m < 1$.

Please help.

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1 Answer 1

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Let $m=1-n$, where $n \in [0,1]$. Then

$(p+q)^m = (p+q)^{1-n} = p (p+q)^{-n} + q (p+q)^{-n} \leq p p^{-n} + q q^{-n} = p^m + q^m$.

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    $\begingroup$ (Conversely, if $m \geq 1$ then $(p+q)^m \geq p^m + q^m$ with the same method.) $\endgroup$
    – sdcvvc
    Dec 23, 2012 at 15:32
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    $\begingroup$ Does this theorem has a name? Is it a type of Jensen's Inequality? $\endgroup$
    – luchonacho
    Mar 9, 2017 at 11:45
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    $\begingroup$ @luchonacho I don't know of a name, and it doesn't seem to be Jensen because it relies only on monotonicity not convexity. $\endgroup$
    – sdcvvc
    Mar 12, 2017 at 15:06
  • $\begingroup$ stupid question I know, but how does this actually prove the statement? $\endgroup$ Nov 17, 2021 at 21:54
  • $\begingroup$ @Onamission The first term in the equation is $(p+q)^m$, the last term is $p^m+q^m$ and they're connected by a string of equalities and inequalities, which proves $(p+q)^m \leq p^m + q^m$. Is something else unclear? $\endgroup$
    – sdcvvc
    Nov 28, 2021 at 15:52

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