<s>[INST] How find the minimum of the $n$ such $99^n+100^n<101^n$ Question:

Find the smallest positive integer $n$ such that
  $$99^n+100^n<101^n$$

My idea: This is equivalent to
$$\left(\dfrac{99}{101}\right)^n+\left(\dfrac{100}{101}\right)^n<1$$
so
$$\left(1-\dfrac{2}{101}\right)^n+\left(1-\dfrac{1}{101}\right)^n<1$$
Use this:
$$(1+x)^n\ge 1+nx,x>-1,n\ge 1$$
so
$$1-\dfrac{2}{101}n+1-\dfrac{1}{101}n<1\Longrightarrow n>\dfrac{101}{3}=33.666$$
But I found $n=34$ is not sufficient. I used a computer to find $n\ge 49$.
I want to see how to find it by hand. Thank you.
 [/INST] First of all, $n \ge 34$ is a necessary (but not sufficient) condition for that inequality to be true. Remember that $1+nx \le c$ does not imply that $(1+x)^n \le c$. 
We can get a better bound by manipulating the inequality as follows: 
$99^n + 100^n < 101^n$
$101^n - 99^n > 100^n$
$\left(\dfrac{101}{100}\right)^n - \left(\dfrac{99}{100}\right)^n > 1$
$\left(1+\dfrac{1}{100}\right)^n - \left(1-\dfrac{1}{100}\right)^n > 1$
Using the binomial theorem, this becomes: 
$2\dbinom{n}{1}\dfrac{1}{100} + 2\dbinom{n}{3}\dfrac{1}{100^3} + 2\dbinom{n}{5}\dfrac{1}{100^5} + \cdots > 1$
All the terms on the left side are positive and first term is $\dfrac{n}{50}$, so the inequality holds for $n \ge 50$. 
This is a sufficient condition, but not a necessary one. For $n = 49$, the first two terms exceed $1$, so the inequality holds for $n = 49$ as well. 
Now, all that remains is to show that the inequality does not hold for $n \le 48$.
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