# Proving inequality question using mathematical induction

I'm having a really hard time on how to prove this using mathematical induction: For all real $$x>-1$$, $$(1+x)^n\geq 1+nx$$.

If $$x>-1$$ then $$1+x>0$$
Base case $$n=1$$: $$(1+x)^1\ge 1+x$$.
Induction Assumption: Assume that works for any $$k\geq 1$$, $$(1+x)^k\geq 1+kx$$
Inductive step: Show that $$(1+x)^{k+1} \geq 1+(k+1)x$$
$$(1+x)^k\geq 1+kx$$
Multiply $$(1+x)$$ on both sides:
$$(1+x)(1+x)^k\ge (1+x)(1+kx)$$
$$(1+x)(1+x)^k\ge 1+kx+x+kx^2$$
$$(1+x)^{k+1}\ge1+(k+1)x+kx^2$$
Since $$kx^2\ge0$$
$$(1+x)^{k+1}\ge1+(k+1)x$$

• This is known as Bernoulli's Inequality and that link contains a proof by innduction. – lulu Oct 4 at 20:37
• Does this help? – Dr. Mathva Oct 4 at 20:37
• Didn't know this was called Bernoulli's Inequality, thanks. I'm looking at some resolutions and trying to understand every step of it. – Luís Fernando Oct 4 at 20:59

$$(1+x)^{k+1}=(1+x)^k(1+x)\geq (1+kx)(1+x)=1+x+kx+kx^2\geq 1+x+kx=$$
$$=1+(k+1)x$$
We just used the inequality $$kx^2\geq 0$$.
• I forgot to ask a step-by-step solution, cause I wasn't seeing the $(1+x)$ multiplication on both sides. I edited my draft, hope it's correct now. – Luís Fernando Oct 4 at 21:25