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

Question

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?

Answer 1

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"

Answer 2

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.

Answer 3

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.

Answer 4

This is not induction but will help$$\begin{align} (1+2x)^n&={\binom n 0}1+{\binom n 1}\cdot2x+{\binom n 2}\cdot(2x)^2+\cdots+{\binom n n}\cdot(2x)^n\\ &=1+2nx+{\binom n 2}\cdot(2x)^2+\cdots+{\binom n n}\cdot(2x)^n\\ \end{align}$$

$$(1+2x)^n\ge1+2nx$$