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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!

Biggly Bobb
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    See https://math.stackexchange.com/q/2268452/42969: It gives $a^n + b^n \ge 2^{n-1}(a+b)^n$. – Martin R Nov 11 '20 at 09:36
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    A simple one would be $k\cdot (\min_i{a_i})^N$. – Milten Nov 11 '20 at 09:36
  • @MartinR that seems perfect! Thanks so much! – Biggly Bobb Nov 11 '20 at 09:50
  • Did you mean $2^{1-n}$ instead of $2^{n-1}$? @MartinR – W. Wongcharoenbhorn Nov 11 '20 at 10:23
  • @W.Wongcharoenbhorn: You are completely right. – I made the same error in the initial version of my answer (where I have fixed it now). – Martin R Nov 11 '20 at 10:26
  • @BigglyBobb: Any feedback on the answer? It would be nice to know if it is of any help. – Martin R Nov 23 '20 at 13:08
  • @MartinR I haven't yet solved the problem I'm working on (which I won't post as it's for a research-type subject), but I've gotten a very similar relation to the relation i used to solve the simpler case for the problem. I suspect that it isn't the inequality which needs a change, but I need to modify the argument to make it work. Thanks so much for the help! – Biggly Bobb Nov 24 '20 at 08:26

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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 $$ N \ge \frac{\log M - \log k}{\log(a_1 + \ldots + a_k) - \log k} \, . $$

Martin R
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