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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.

1 Answers 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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    @luchonacho I don't know of a name, and it doesn't seem to be Jensen because it relies only on monotonicity not convexity.2017-03-12