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

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

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

2. Hello,

As you said, we have :

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

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

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

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

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

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

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

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

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

Looks clear to you ?

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