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 Answer 1


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

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

  • 1
    $\begingroup$ (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
  • 2
    $\begingroup$ Does this theorem has a name? Is it a type of Jensen's Inequality? $\endgroup$
    – luchonacho
    Mar 9, 2017 at 11:45
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.