Hi guys

Need help with this proof buy induction.

SOLUTION:Let $x\in REAL$ with $x\geq-1$. Prove that $(1+x)^n\geq1+nx$ for every integer $n\geq1$

Let $P(n)$ be the statement that $(1+x)^n\geq1+nx$

anchor step: $P(1)$obviously trueinduction step:

suppose $x\geq1$ is an integer and that $P(n)$ is true

since $x\geq-1$ we have $1+x\geq0$ and so using $P(n)$ gives

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

$\geq (1+x)(1+nx)$

$=1+x+nx+nx^2=1+(n+1)x+nx^2 \geq 1+(n+1)x$

so we have deduced $P(n+1)$ . Hence $P(n)$ holds for all $n \geq 1$ .

i understand $(1+x)^{n+1}=(1+x)\cdot(1+x)^n$ by law of exponentials

ok here's the problem.

but then this : $\geq (1+x)(1+nx)$ ???? I mean where does this come from: $ (1+x)(1+nx)$

obviously in the hypothesis we have $(1+x)^n\geq1+nx$ ... But how to get from $(1+x)^n\geq1+nx$ to $(1+x)^{n+1} \geq (1+x)(1+nx)$ ?? I tried to sub-in $n+1$ and it didnt work..

$(1+(n+1)x) = nx + x + 1 = ????$

How to get from here to $(1+x)(1+nx)$ ????

Obviously im missing something

Thank u guys