## Bernoullis inequality

Show that for all $$n\in \mathbb N$$ and $$x\in \mathbb R,\ x> -1$$ it is true that $$(1+x)^n \ge 1+nx$$.
For $$n=1$$ certainly $$1+x\ge 1+x$$.
Given that the statement is true for $$n$$, we show that it holds for $$n+1$$:
$$(1+x)^{n+1}=(1+x)^n(1+x)\ge (1+nx)(1+x)=1+(n+1)x+nx^2\ge 1+(n+1)x$$.
Notice that in the first inequality we used the assumption that $$1+x$$ is non-negative.