how to prove the inequality (1-x^2)^n>=1-nx^2?

**zzzhhh** · September 27th 2009, 05:42 AM

Let $x\in [0,1]$ and $n$ be any positive integer, prove $(1-x^2)^n\geq1-nx^2$ .
It can be easily proved by induction on $n$ , but I would like to use binomial theorem to prove it. By binomial theorem, $(1-x^2)^n=1-nx^2+...$ . The proof is completed if the remaining "..." part in the above expansion can be proved to be $\geq0$ , but I have no idea how to prove it. Could anyone help me with this problem? Thanks!

**Moo** · September 27th 2009, 09:25 AM

Hello,

As you said, we have :

$(1-x^2)^n=\sum_{k=0}^n {n\choose k} (-x^2)^{k}=1-nx^2+\sum_{k=2}^n {n\choose k} (-x^2)^{k}$

Now let's consider $S=\sum_{k=2}^n {n\choose k} (-x^2)^{k}$
Try to make the sum start by 0, but not by adding terms, otherwise we'd be back to the beginning.
Instead, let $j=k-2$

Thus $S=\sum_{j=0}^{n-2} {n\choose j+2} (-x^2)^{j+2}$

$S=\sum_{j=0}^{n-2} \frac{n!}{(j+2)!(n-j-2)!} \cdot (-x^2)^{j+2}$

Now factor out some things in order to get a binomial expansion :

$S=n(n-1)(-x^2)^2\sum_{j=0}^{n-2} \frac{(n-2)!}{j!(n-2-j)!}\cdot (-x^2)^j$

$S=n(n-1)x^4\sum_{j=0}^{n-2} {n-2 \choose j} (-x^2)^j$

Which is exactly $n(n-1)x^4(1-x^2)^{n-2}$

And this is positive, since $1-x^2\geq 0$

Looks clear to you ?

**zzzhhh** · September 29th 2009, 01:04 AM

Thank you but, during the factoring, how to turn the $(j+2)!$ in the denominator into $j!$ ?

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