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

Edit: (Solution with the help of comments and answer below)

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 $kx^2\ge0$
$(1+x)^{k+1}\ge1+(k+1)x$

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

$(1+x)^{k+1}=(1+x)^k(1+x)\geq (1+kx)(1+x)=1+x+kx+kx^2\geq 1+x+kx=$

$=1+(k+1)x$

We just used the inequality $kx^2\geq 0$.

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  • $\begingroup$ 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. $\endgroup$ – Luís Fernando Oct 4 at 21:25

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