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.

The proof by induction is natural and fairly straightforward, but it’s worth pointing out that induction isn’t actually needed for this result if one has the binomial theorem at hand:
Corrected:
$$\begin{align*}
(1-x)^n+(1+x)^n &= \sum_{k=0}^n\binom{n}k (-1)^kx^k + \sum_{k=0}^n \binom{n}k x^k\\
&= \sum_{k=0}^n\binom{n}k \left((-1)^k+1\right)x^k\\
&= 2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}\\
&< 2\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}\tag{1}\\
&= 2\cdot 2^{n-1}\tag{2}\\
&= 2^n,
\end{align*}$$ where the inequality in $(1)$ holds because $|x|< 1$, and $(2)$ holds for $n>0$ because $\sum\limits_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}$, the number of subsets of $[n]=\{1,\dots,n\}$ of even cardinality, is equal to the number of subsets of $[n]$ of odd cardinality.
