Prove that $(1+x)^n ≥1+nx$ for all $x>-1$ and $n=1,2,\ldots$

Question

Prove that for every real number $x > −1$ and every $n = 1,2,\ldots,$ $$(1+x)^n ≥1+nx.$$

I don't know where to begin so I haven't tried anything.

Answer 1

this the so called Bernoulli inequality, $(1+x)^n\geq 1+nx$ we can prove it by induction

Answer 2

let $f(x) = (1+x)^n - 1 - nx$. Can you find what the variations of $f$ are?

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Alternative:

let $g(x) = (1+x)^n$. Then $n\ge 1\implies g$ is convex. Let $x>-1$. Using the mean value theorem, $$ g(x) = g(0) + xg'(0) + g''(c) $$for a certain $c$ between $0$ and $x$. As $g''\ge 0$, $$ g(x) \ge g(0) + xg'(0) = 1+nx $$