2
$\begingroup$

I know this isn't a very typical question, but i was wondering if anyone knows any good lower bounds for $a^n + b^n$. I'm looking for something akin to $a^n + b^n \leq (a + b)^n$ for $n \geq 1$.

The motivation is that I'm trying to find some nice upper bounds on $N$ for the minimum value of $N$ when $a_1^N + a_2^N + \dots + a_k^N \geq M$, for some given $1 \leq a_1, \dots, a_k \leq 2$ and $M > 0$. I don't need this to be particularly tight, but I was thinking if I could write this as some expression $a_1^N + a_2^N + \dots + a_k^N \geq f(a_1, a_2, \dots, a_k)^N \geq M$, then if $f$ is nice enough the $\log$ upper bound on $N$ should be good enough for my purposes.

I've tried using the AM-GM to get $a^n + b^n \geq nab$, but as that doesn't give a logarthmic bound it's not good enough for my purposes. I've scrolled through a bunch of other lists of inequalities and I can't find anything else that seems to work.

Does anyone have any ideas? Either for my original question or for the motivation? Thanks for the help!

$\endgroup$
7
  • 2
    $\begingroup$ See math.stackexchange.com/q/2268452/42969: It gives $a^n + b^n \ge 2^{n-1}(a+b)^n$. $\endgroup$
    – Martin R
    Nov 11 '20 at 9:36
  • 1
    $\begingroup$ A simple one would be $k\cdot (\min_i\{a_i\})^N$. $\endgroup$
    – Milten
    Nov 11 '20 at 9:36
  • $\begingroup$ @MartinR that seems perfect! Thanks so much! $\endgroup$ Nov 11 '20 at 9:50
  • $\begingroup$ Did you mean $2^{1-n}$ instead of $2^{n-1}$? @MartinR $\endgroup$ Nov 11 '20 at 10:23
  • $\begingroup$ @W.Wongcharoenbhorn: You are completely right. – I made the same error in the initial version of my answer (where I have fixed it now). $\endgroup$
    – Martin R
    Nov 11 '20 at 10:26
2
$\begingroup$

With respect to the original problem:

If all $a_j$ are equal to one then $a_1^N + a_2^N + \dots + a_k^N \ge M$ is equivalent to $k \ge M$, independently of $N$.

Otherwise Jensen's inequality for the convex function $t \to t^N$ (see for example Prove inequality using Jensen's inequality) gives the following estimate: $$ a_1^N + \ldots + a_k^N \ge k^{1-N}(a_1 + \ldots + a_k)^N $$ and that is $\ge M$ if $$ \frac{(a_1 + \ldots + a_k)^N}{k^N} \ge \frac M k $$ or $$ N \ge \frac{\log M - \log k}{\log(a_1 + \ldots + a_k) - \log k} \, . $$

$\endgroup$

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.