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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 and $0.

Could you tell me please, how should I move on. And do I need a binomial theorem here?

Thank you in advance.

5 Answers 5

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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.

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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?

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    @ So, it's like we are going from the original inequality to the IS, right? If I multiply the first inequality, then I get: (1-x)^{n+1}+(1+x)^{n+1} + 2((1-x)^{n}+(1+x)^{n})<2\cdot 2^{n} Second part is positive, as $(1-x)^{n}$ positive, and $(1+x)^{n}$, so it means, that if the whole left hand side is smaller than right hand side and both summands are positive, the both of them separately are especially smaller than the right hand side. And so we get: (1-x)^{n+1}+(1+x)^{n+1} < 2\cdot 2^{n} = 2^{n+1} Right?2011-10-19
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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.

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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

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Hint: For the induction step, use the fact that

$(1-x)^{n+1}+(1+x)^{n+1}$ is equal to $[(1-x)^n+(1+x)^n][(1-x)+(1+x)]-[(1-x)(1+x)^n+(1+x)(1-x)^n].$

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    Your hint overshoots the writeable area. Can you split it into two lines?2011-10-18