In the book$^1$ that I am reading, the author dubs an $n$-digit positive number a Smallbrain number if it is equal to the sum of the $n$th powers of its digits, with $371 = 3^3 + 7^3 + 1^3$ given as an example. The question is to prove that there are no Smallbrain numbers with 1000 digits.
The problem is that for $n = 1000$, the sum of the $n$th powers of the digits is at most $1000\cdot9^{1000}$, which is strictly less than the least number with 1000 digits, namely $10^{999}$. I showed this as follows:
$$(*) \iff 9^{1000} < 10^{996}$$
$$\impliedby 9^{1000} < 10^{990}$$
$$\impliedby 9^{10} < \bigg(1+\frac{1}{9}\bigg)^{990}$$
$$\impliedby 9^{10} < 2^{110}$$
using that $\bigg(1+\frac{1}{n}\bigg)^{n} \ge 2.$
This question, however, is printed immediately the after a chapter explaining the AM-GM inequality and Cauchy's inequality for $n = 2$. Although I realise that the fact I used can be derived from AM-GM, I doubt that this was expected for the answer. Is there are a more direct way to prove $(*)$ using the two inequalities I mention?
1: A Concise Introduction to Pure Mathematics, Liebeck.
Best Answer
Weighted AM-GM can solve this pretty easily: $$9^{1000} < 10^{996} \iff 9^{250}<10^{249} \iff 9 <\left(\frac{10}{9}\right)^{249} \iff 9^{\frac{1}{249}} <\frac{10}{9}$$ Note that by AM-GM on $9,1$ with weights $1, 248$ respectively, we get $$9^{\frac{1}{249}}<\frac{9+248}{249}<\frac{10}{9}$$ The last inequality can be easily verified by cross multiplication.