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If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$.

Is there a way to prove it without using the Binomial Theorem? Is it possible to use the Bernoulli's Inequality to prove it? If yes, please show me.

I tried to prove it by induction on $n$, where $x>0$ is fixed. But it doesn't let me reach the conclusion. Here's what I did:

Fix $x>0$. We instead prove that \begin{equation*} \frac{2(1+x)^n}{n(n-1)}> x^2 \end{equation*} for all $n>1$. Clearly, this is true for $n=2$. Assume this statement holds for $n=k$, for some $k\geq 2$. We have \begin{equation*} \frac{2(1+x)^{k+1}}{(k+1)k}=\frac{2(1+x)^k}{k(k-1)}\frac{(x+1)(k-1)}{k+1}>\frac{(x+1)(k-1)}{k+1}x^2. \end{equation*} Since $y\mapsto \frac{y-1}{y+1}$ is increasing on $\mathbb{R}\setminus \{-1\}$, we get $\frac{y-1}{y+1}\geq \frac{1}{3}$ for all $y\geq 2$. Therefore, \begin{equation*} \frac{(x+1)(k-1)}{k+1}x^2\geq \frac{1}{3}(x+1)x^2>\frac{1}{3}x^2. \end{equation*}

As you can see, I can not reach $>x^2$ instead of $>\frac{1}{3}x^2$.

UnknownW
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  • If you can get $k = 3$, then you can use the fact that $\frac{k-1}{k+1}\geq \frac{1}{2}$ for all $k\geq 3$. – Michael L. Sep 16 '18 at 22:17
  • You started by showing the validity for $k=2$, Thus, it's better to say $\frac{k-1}{k+1}\ge\frac12,\colon k\ge3$. – Hazem Orabi Sep 16 '18 at 22:27

2 Answers2

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Let, $f(x)=(1+x)^n-\frac{1}{2} n(n-1) x^2.$

So, $f'(x)=n(1+x)^{n-1}-n(n-1)x>0 \forall n>1$

Since,for $n>1$ and $x\ge 0,$ by A.M.-G.M. inequality we have, $$\frac{1.(1+x)^{n-1}+(n-2).1}{n-1}>(1+x)$$

$$\implies (1+x)^{n-1}>(n-1)x$$

Therefore, $f$ is strictly increasing. Hence we have, $$f(x)>f(0)\implies (1+x)^n>\frac{1}{2}n(n-1) x^2$$

SOUL
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As an alternative we can proceed by induction for the stronger

$$(1+x)^{n}>1+nx+\frac12n(n-1)x^2>\frac12n(n-1)x^2$$

that is for the induction step

$$(1+x)^{n+1}=(1+x)(1+x)^{n}\stackrel{Ind.Hyp.}>1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3>$$$$\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$

therefore we need to prove that

$$1+nx+\frac12n(n-1)x^2+x+nx^2+\frac12n(n-1)x^3\stackrel{?}>1+(n+1)x+\frac12(n+1)nx^2$$

which is true indeed

$$\frac12n(n-1)x^2+nx^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$

$$\frac12\left(n(n-1)+2n\right)x^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$

$$\frac12\left(n^2+n\right)x^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$

$$\frac12(n+1)nx^2+\frac12n(n-1)x^3\stackrel{?}>\frac12(n+1)nx^2$$

$$\frac12n(n-1)x^3>0$$

for $n>1$.

user
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