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.
 A: You "basis" proof is upside down: you should start with what is known and work towards what you want to prove.
Can you see $(1-x)^n$ and $(1+x)^n$ are each positive if $|x| < 1$? And $(1-x)$ and $(1+x)$ are each less than $2$?  
So $(1-x)^{n+1}+(1+x)^{n+1} < 2(1-x)^{n}+2(1+x)^{n}$ and you should be able to complete the induction.
A: Hint:  For the induction step, you assume $(1-x)^n+(1+x)^n<2^n$.  You want to prove that $(1-x)^{n+1}+(1+x)^{n+1}<2^{n+1}$  If you multiply both sides of the first inequality by $(1-x)+(1+x)=2$ what happens?
A: 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.
A: 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 
A: Hint: For the induction step, use the fact that 
$$(1-x)^{n+1}+(1+x)^{n+1}$$ 
is equal to
$$[(1-x)^n+(1+x)^n][(1-x)+(1+x)]-[(1-x)(1+x)^n+(1+x)(1-x)^n].$$
