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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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3 Answers3

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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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    @luchonacho I recently realized this is indeed closely related to the concavity of $x^a$ for $a<1$. The other user is not quite right in saying it relies only on monotonicity, since it also relies on the factoring property used in the proof to "steal" one power of $p, q$ away. The inequality in this MSE question is an example of a much more general concept called convex order of random variables (which also covers Jensen's inequality), which I explain here https://math.stackexchange.com/a/5041495/405572 – D.R. Mar 03 '25 at 06:46
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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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There's a more general way to look at this.

Let $0< m< 1$. Consider the function $f(x)=x^m$.

It is concave on $[0,\infty)$ and $f(0)\geq 0$. It has been shown here that such a function $f$ is sub-additive i.e., $f(p+q)\leq f(p)+f(q)$ for all $p,q\in [0,\infty)$.

[For those unversed, concave means that the graph of the function is above its secant line. I got to know that $f$ is concave using the second derivative test. We see that $f''(x)=\frac{m(m-1)}{2}x^{m-2}$ is negative on $(0,\infty)$.]

The case $m=1$ and $m=0$ can be handled easily by simple substitution.

Thus, $(p+q)^m\leq p^m+q^m$ for all $p,q>0$ and $0\leq m\leq 1$. $\blacksquare$

Nothing special
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  • thanks for linking the other answer. It helped me realize that there is an even more general way to look at this, namely a notion called convex ordering of random variables, which in particular proves the subadditivity property you quote, and also for instance the famous Jensen inequality. I explain it here https://math.stackexchange.com/a/5041495/405572 – D.R. Mar 03 '25 at 06:47