Proving $(1-x)^n \geq 1 - nx $

Question

$(1-x)^n \geq 1 - nx\,\, $

If i expand the left side of the inequality with the binomianl coefficient formula I obtain:

$1-nx + {n \choose 2}x^2 - {n \choose 3}x^3 ... $ now I see where the $1-nx$ in the inequality came from, but how do I prove that $$ {n \choose 2}x^2 - {n \choose 3}x^3 ...> 0 $$

Answer 1

Well inequality holds if $x\in[0,1]$ and $n\geq 1.$ A typical proof:

Set $f(x)=(1-x)^n+nx-1$ then $f'(x)=-n(1-x)^{n-1}+n=n[1-(1-x)^n]\ge 0.$ So $f $ is increasing , note that $f(0)=0$ therefore $f(x)\geq 0.$

Answer 2

By induction over n, assuming $x\in [0,1]$.

• $n=1$

We have $1-x = 1-x$;

• Suppose that for $n=k$ is valid. Then $(1-x)^{k} \geq 1-kx \Rightarrow (1-x)^{k+1} \geq (1-kx)(1-x) = 1 -x-kx+kx^2 \\ = 1 -(k+1)x + kx^2 \geq 1-(k+1)x \Rightarrow (1-x)^{k+1} \geq 1-(k+1)x$

for $k>0$ and $x\in[0,1]$.

There it is.

As an exercise check:

i.$(1+x)^{n}\geq 1+nx$ when $x\geq -1$ and $\forall n \in \mathbb{N}$;

ii.$(1+x)^{2n}>1+2n$ , $\forall x \in \mathbb{R - \lbrace 0 \rbrace}$.