Prove $x^n\ge 1+n(x-1) \quad \forall x \ge 1, \forall n \in N$

Question

My attemp:

using the induction principle :

if $n=1$ I have an identity $x\ge x$.I suppose the principle is true for $n$ and I want to prove it for $n+1$

$x^{n+1} = x^n \cdot x\ge [1+n(x-1)]\cdot x$ but how can I obtained $1+(n+1)(x-1)$?

Answer 1

Here is a one liner,

$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+1)\geq (x-1) \underbrace{(1+\cdots +1)}_\text{n of them}=n(x-1)$

Answer 2

Note that $x+nx(x-1)-1-(n+1)(x-1)=x-1+(x-1)(nx-n-1)=n(x-1)^2\ge0$.

Answer 3

Note that, $(x-1)^2\geq 0\implies x^2-x\geq x-1\implies nx^2-nx\geq nx-n$. Hence,

$$x\big[1+n(x-1)\big]=x+nx(x-1)=x+nx^2-nx\geq nx-n+x$$$$= 1+nx-n+x-1=1+(n+1)(x-1).$$

Answer 4

Binomial expansion

$x \ge 1$;

$x^n=(1+(x-1))^n=$

$1+n(x-1)+$

$\displaystyle{\sum_{k=2}^n}\binom{n}{k}1^{n-k}(x-1)^{k} \ge$

$1+n(x-1).$