Prove that $(p+q)^m \leq p^m+q^m$

Question

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.

sdcvvc · Accepted Answer · 2012-12-23 15:28:44Z

45

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

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

answered Dec 23 '12 at 15:28

sdcvvc

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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 '12 at 15:32
2

$\begingroup$ Does this theorem has a name? Is it a type of Jensen's Inequality? $\endgroup$ – luchonacho Mar 9 '17 at 11:45
$\begingroup$ @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 '17 at 15:06

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