How can I prove this $(1 + x)^n \geq n\cdot x + 1$

Question

A theorem says:

$\forall x > -1$ and $\forall n \in \mathbb{N}$:

$(1 + x)^n \geq n\cdot x + 1$

Base:

$n = 0$

$(1 + x)^0 \geq 0\cdot x + 1$

$1 \geq 1$ [ok]

Hypothesis:

Suppose $(1 + x)^n \geq n\cdot x + 1$ is true for an arbitrary $n$.

Inductive step:

Let's prove $(1 + x)^{n+1} \geq (n+1)\cdot x + 1$ is also true.

$(1 + x)^{n} \cdot (1 + x) \geq n\cdot x + 1 + x$

I know $(1 + x)^{n}$ is $\geq n\cdot x + 1$, by inductive hypothesis, but I cannot figure out how to simplify the expression above further, it seems I am very close to the solution, but (maybe) I am too stupid without imagination.

Answer 1

You know that $(1+x)^n\ge nx+1$ by the inductive hypothesis. Let's multiply both sides by the positive quantity $1+x$. This will maintain the inequality. We get $$(1+x)^{n+1}\ge (nx+1)(1+x)=nx+1+nx^2+x=(n+1)x+1+nx^2\ge (n+1)x+1$$

Answer 2

The inductive step is equivalent to $$(1+x)^n(1+x)=(1+x)^n+x(1+x)^n\ge n\cdot x+1 +x,$$ which, given the hypothesis, is implied by $$x(1+x)^n\ge x,\tag{$\star$}$$ true for all $x\ge-2$. Indeed, since $(\star)$ trivially holds for $x=0$ , divide both LHS and RHS by $x$ to get $$ \left\{ \begin{array}{c} (1+x)^n\ge 1 & x>0 \\ (1+x)^n\le 1 & x<0 \end{array} \right. , $$

which is clearly always true for $x\ge-2$.

Answer 3

$$(1+x)^{n+1} = (1+x)^n(1+x) \ge (1+nx)(1+x)=1+(n+1)x+nx^2 \ge1+(n+1)x $$ since $nx^2\ge0$.

Answer 4

\begin{align} (1+x)^{n+1}&=(1+x)(1+x)^n \\ &\ge (1+x) (nx+1) \\ &=nx+1+nx^2+x \\ &=(n+1)x+1+nx^2 \\ &\ge(n+1)x+1 \end{align}

Answer 5

It's Bernoulli's inequality. You just have to start from the inductive hypothesis: $\,\ (1+x)^n \ge 1+nx$, and multiply both members of this inequality by $1+x$ (which $>0$). You obtain: $$ (1+x)^{n+1}\ge 1+nx+x+nx^2=1+(n+1)x+nx^2\ge 1+(n+1)x. $$

You may try to prove the following generalisation, which may be useful: if $a_1, \dots, a_n$ are numbers all $>-1$, then $$(1+a_1)\dotsm\,(1+a_n)\ge 1+a_1+\dots+a_n.$$