Use Induction to prove: $(1+2x)^n \geq 1+2nx$ Show by induction that:

for all $x>0$ that $(1+2x)^n \geq 1+2nx$

So far I have:

for $n=1 \rightarrow (1+2x)^1 \geq 1+2x$. True!
for $n=k+1 \rightarrow (1+2x)^{k+1} \geq 1+2(k+1)x$
= $(1+2x)^k (1+2x) \geq 1+2xk+2x $

What is te next step to show this is true?
 A: Taking  it from where you left it:
$$(1+2x)^k(1+2x)\stackrel{\text{Ind. hypothesis}}\ge(1+2kx)(1+2x)$$
So it is enough to show
$$(1+2kx)(1+2x)\ge1+2(k+1)x$$
and this is true iff (opening parentheses)
$$1+2x+2kx+4kx^2\ge1+2kx+2x\iff4kx^2\ge0$$
and since the last inequality is trivial we're done.
You may want to google "Bernoulli inequality"
A: Hint: Use the Binomial Theorem to compare the terms.
[edit] In John Zhangs answer, the finding that $(1+2x)^k$ was greater or equal to $(1+2xk)$ can be found easily with the Binomial Theorem, so its a good way to expand any part of any binomial power.
A: for $n=1 \rightarrow (1+2x)^1 \geq 1+2x$. True!
for $n=k+1 \rightarrow (1+2x)^{k+1} =(1+2x)^k (1+2x) \geq (1+2xk)(1+2x)> 1+2(k+1)x$ 
here we use induction hypothesis.
A: This is not induction but will help$$\begin{align}
(1+2x)^n&={\binom n 0}1+{\binom n 1}\cdot2x+{\binom n 2}\cdot(2x)^2+\cdots+{\binom n n}\cdot(2x)^n\\
&=1+2nx+{\binom n 2}\cdot(2x)^2+\cdots+{\binom n n}\cdot(2x)^n\\
\end{align}$$
$$(1+2x)^n\ge1+2nx$$
