Prove that $(p+q)^m \leq p^m+q^m$
Solution 1:
Let $m=1-n$, where $n \in [0,1]$. Then
$(p+q)^m = (p+q)^{1-n} = p (p+q)^{-n} + q (p+q)^{-n} \leq p p^{-n} + q q^{-n} = p^m + q^m$.
Prove that $(p+q)^m \leq p^m+q^m$
Solution 1:
Let $m=1-n$, where $n \in [0,1]$. Then
$(p+q)^m = (p+q)^{1-n} = p (p+q)^{-n} + q (p+q)^{-n} \leq p p^{-n} + q q^{-n} = p^m + q^m$.