Prove an inequality by Induction: $(1-x)^n + (1+x)^n < 2^n$ Could you give me some hints, please, to the following problem.
Given $x \in \mathbb{R}$ such that $|x| < 1$. Prove by induction the following inequality for all $n \geq 2$:
$$(1-x)^n + (1+x)^n < 2^n$$
$1$ Basis: 
$$n=2$$
$$(1-x)^2 + (1+x)^2 < 2^2$$
$$(1-2x+x^2) + (1+2x+x^2) < 2^2$$
$$2+2x^2 < 2^2$$
$$2(1+x^2) < 2^2$$
$$1+x^2 < 2$$
$$x^2 < 1 \implies |x| < 1$$
$2$ Induction Step: $n \rightarrow n+1$
$$(1-x)^{n+1} + (1+x)^{n+1} < 2^{n+1}$$
$$(1-x)(1-x)^n + (1+x)(1+x)^n < 2·2^n$$
I tried to split it into $3$ cases: $x=0$ (then it's true), $-1<x<0$ and $0<x<1$.
Could you tell me please, how should I move on. And do I need a binomial theorem here?
Thank you in advance.

Just show that $a^n + b^n \le (a+b)^n$ 
( i guess you can use induction here too because if $p>0$ and $q>0$ we have $p+q>0$)
since $a,b$ are positive in the case
after expansion you get 
$\sum_{i=1}^{n-1} \frac{n!}{i!(n-i)!} a^i b^{n-i}>0$  Which is obviously true 
