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 $$ {n \choose 2}x^2 - {n \choose 3}x^3 ...> 0 $$

For $x=0$ you get $1>1$ for any $n$, so the statement is false — ziggurism, Sep 08 '14 at 23:29
@DiegoMath: This is true for all $x \leq 1$ and positive integers $n$. — Paramanand Singh, Sep 09 '14 at 08:02
@ParamanandSingh the question was modified, now it's correct. — DiegoMath, Sep 10 '14 at 01:51
@DiegoMath Wikipedia presents this Alternative form. And see this answer. — , Apr 23 '23 at 21:14

score 3 · Accepted Answer · answered Sep 08 '14 at 23:42

3

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 $f(x)\geq 0.$

answered Sep 08 '14 at 23:42

BigM

4,064

Aaron Maroja · Answer 2 · 2014-09-09T00:42:53.003

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

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

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