# Nice lower bounds for $a^n + b^n$ for $a,b,n \geq 1$?

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!

• See math.stackexchange.com/q/2268452/42969: It gives $a^n + b^n \ge 2^{n-1}(a+b)^n$. Nov 11 '20 at 9:36
• A simple one would be $k\cdot (\min_i\{a_i\})^N$. Nov 11 '20 at 9:36
• @MartinR that seems perfect! Thanks so much! Nov 11 '20 at 9:50
• Did you mean $2^{1-n}$ instead of $2^{n-1}$? @MartinR 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). Nov 11 '20 at 10:26

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