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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.

Martin Sleziak
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Argha
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2 Answers2

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

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

sdcvvc
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    (Conversely, if $m \geq 1$ then $(p+q)^m \geq p^m + q^m$ with the same method.) – sdcvvc Dec 23 '12 at 15:32
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    Does this theorem has a name? Is it a type of Jensen's Inequality? – luchonacho Mar 09 '17 at 11:45
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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. – sdcvvc Mar 12 '17 at 15:06
  • stupid question I know, but how does this actually prove the statement? – On a mission Nov 17 '21 at 21:54
  • @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? – sdcvvc Nov 28 '21 at 15:52
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Here is an alternative proof. Dividing by $(p+q)^m$, it is sufficient to show that $$ 1 < x^m + (1-x)^m$$ when $0<x<1$. But it is easy to see that $x < x^m$ when $m<1$ and $0<x<1$, from which the inequality follows directly: $$ 1 = x + (1-x) < x^m + (1-x)^m.$$ When $m>1$, $x>x^m$, and the inequality is reversed. Equality holds trivially if $m=1$. The proof generalizes in an obvious way to an arbitrary number of summands.

To see that $x < x^m$ if $m <1$, and that $x > x^m$ if $m>1$, take logarithms of both sides and divide by $\log x<0$.

Per Mattson
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