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.
 A: 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 $n \le 48$.
A: Considering the problem from a purely algebraic point of view : if you plot the function $y=\log(99^n+100^n)-\log(101^n)$ (which is basically a straight line since $\frac {d^2y}{dn^2}$ is $0.0000252523$ for $n=0$ and $0.0000198134$ for $n=100$, you would see that $y=0$ if $n=48.2275$.
So the inequality holds for $n \geq 49$.
If you perform one iteration of Halley method starting at $n=0$, it predicts an underestimate of the solution at $n=48.0917$.
A: In view of JimmyK4542's answer it remains to show that
$$\left({99\over101}\right)^{48}+\left({100\over101}\right)^{48}>1\ ,$$
which is the same thing as
$$\left(1+{2\over99}\right)^{-48}+\left(1+{1\over100}\right)^{-48}>1\ .$$
Now
$$(1+x)^{-n}=\sum_{k=0}^\infty{n+k-1\choose k}(-x)^k\ .$$
When $0<x<{k+1\over n+k}$ the terms of this alternating series are decreasing in absolute value after the $k$th term. In the following $x<{1\over 49}$, $n=48$, and $k=5$; therefore we are on the save side when we write
$$(1+x)^{-48}>1 -48x+ 1176x^2 -19600x^3+ 249900x^4-2598960x^5\qquad(0<x<{1\over 49})\ .$$
Putting $x:={2\over99}$ and $x={1\over100}$ here produces
$$\left(1+{2\over99}\right)^{-48}+\left(1+{1\over100}\right)^{-48}>{12404629614100147\over12382682941406250}>1\ .$$
