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I apologize if the title is confusing, but I essentially just want to know if there exists an accepted, rigorous proof of the inequality discussed in this question.

It seems as though it would have to be true that given $k \geq 1$ and $a_{i} \geq 0$, we'll have that:

$\sum_{i=0}^{n}a_{i}^{k} \leq (\sum_{i=0}^{n}a_{i})^{k}$,

but I'm having trouble coming up with a definitive proof myself that works for any $k \geq 1$, regardless of whether it's an integer or not, and I haven't been able to find any published proofs for this inequality despite spending a significant amount of time searching for one.

The accepted answer to the question I linked above seems completely reasonable, but it's certainly not rigorous. Add to this the fact that I can't find a rigorous proof of this inequality, or even any discussion of it whatsoever for that matter, anywhere else on the internet, despite spending over an hour reading through the articles for every significant and/or named inequality on wolfram mathworld and wikipedia, and my confidence that this inequality is in fact valid is not very high.

Can anyone provide any insight into this, or point me to a source that says something definitive about this inequality one way or the other? I'm especially interested in the cases for which $k$ is not an integer.

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    $\begingroup$ If you expand $(\sum a_i)^k$ you get a lot of terms, including all the $a_i^k$, and others which are definitely nonnegative. $\endgroup$ – Lord Shark the Unknown Apr 13 '18 at 17:14
  • $\begingroup$ For two variables we have $$\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '18 at 17:15
  • $\begingroup$ @LordSharktheUnknown The OP specifically asks about the case where $k$ is not an integer. $\endgroup$ – saulspatz Apr 13 '18 at 17:17
  • $\begingroup$ @Dr.SonnhardGraubner That comes from Holder's Inequality, doesn't it? I tried to see if I could get that into a form that would be applicable for the case I'm interested in but it doesn't seem like it's able to really give me any definitive insight either way. $\endgroup$ – bones_mccoy Apr 13 '18 at 17:25
  • $\begingroup$ What you're looking for is a version of Minkowski's inequality $\endgroup$ – Omnomnomnom Apr 13 '18 at 17:31
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Let $\lambda = \sum\limits_{i=0}^n a_i$. If $\lambda = 0$, all $a_i = 0$ and the inequality is trivially true.

WOLOG, let's assume some $a_i > 0 \implies \lambda > 0$.

For any $i = 0, 1,\ldots, n$, let $\displaystyle\;b_i = \frac{a_i}{\lambda}$, we have $\;b_i \in [0,1] \implies b_i^k \le b_i$.

Together with $\sum\limits_{i=0}^n b_i = 1$, we find

$$\sum_{i=0}^n a_i^k = \lambda^k \sum_{i=0}^n b_i^k \le \lambda^k \sum_{i=0}^n b_i = \lambda^k = \left(\sum_{i=0}^n a_i\right)^k$$

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If you write $$f(x_1,x_2,\dots,x_n)=\left(\sum{x_i}\right)^k-\sum{x_i^k},$$ then $f(0,0,...,0)=0$ and it's easy to show that all the first-order partial derivatives $\frac{\partial f}{\partial x_1}$ are strictly positive when $k>1.$ Indeed,$$ \frac{\partial f}{\partial x_1}=k\left(\left(\sum{x_i}\right)^{k-1}-x_i^{k-1}\right)>0$$ when any of the $x_i>0$.

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If you can prove $a^k+b^k\le(a+b)^k$ for $a$, $b>0$, then the general case follows by induction. This is the same as $t^k+(1-t)^k\le1$ for $0<t<1$, taking $t=a/(a+b)$, By calculus, $t^k+(1-t)^k$ has only one extremum in $(0,1)$, at $t=1/2$, and that's a minimum when $k>1$. So on $[0,1]$, $t^k+(1-t)^k\le1$.

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