24
$\begingroup$

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.

$\endgroup$
38
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ (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
  • 1
    $\begingroup$ 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

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.