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# Prove that $(1+x)^n \geq (1+nx)$, for all natural number $n$, where $x > -1$.

Let $P(n)$ be the given statement,
i.e., $P(n) : (1+x)^n \geq (1+nx)$, for $x > -1$.
We note that $P(n)$ is true when $n=1$, since $(1+x) \geq (1+x)$ for $x > -1$.
Assume that
$P(k) : (1+x)^k \geq (1+kx), x > -1$ is true.-----------(1)
We want to prove that $P(k+1)$ is true for $x > -1$ whenever $P(k)$ is true.--------(2)
Consider the identity
$(1+x)^{k+1}=(1+x)^k(1+x)$
Given that $x > -1$ so $(1+x) > 0$.
Therefore, by using $(1+x)^k \geq (1+kx)$ we have
$\qquad (1+x)^{k+1} \geq (1+kx)(1+x)$
i.e., $(1+x)^{k+1} \geq (1+x+kx+kx^2).$------------(3)
Here $k$ is a natural number and $x^2 \geq 0$ so that $kx^2 \geq 0$. Therefore
$(1+x+kx+kx^2) \geq (1+x+kx),$
and so we obtain
$\qquad (1+x)^{k+1} \geq (1+x+kx)$
i.e., $(1+x)^{k+1} \geq [1+(1+k)x]$
Thus, the statement in (2) is established. Hence by the principle of mathematical induction, $P(n)$ is true for all natural numbers.

edited May 3, 2014