Proving an inequality by mathematical induction

Question

I'm trying to solve a problem with inequalities using mathematical induction but I am stuck halfway through the process. The problem: Use mathematical induction to establish the inequality - $(1 + \frac{1}{2})^n \ge 1 + \frac{n}{2}$ for n $\in \mathbb{N}$

Steps

1) $n = 1$, $(1 + \frac{1}{2})^1 \ge 1 + \frac{1}{2}$ is TRUE

2) $n = k$, assume that $(1 + \frac{1}{2})^k \ge 1 + \frac{k}{2}$ for n $\in \mathbb{N}$

3) Show the statement is true for $k + 1$

$(1 + \frac{1}{2})^{k+1}$ = $(1 + \frac{1}{2})^k * (1 + \frac{1}{2})$

$\ge$ $(1 + \frac{k}{2}) * (1 + \frac{1}{2})$ - using the assumption in step $2$

My question is, how do I continue this problem? Or did I go wrong somewhere? I just can't figure out what the next step is.

Answer 1

Continue with:

$(1 + \frac{k}{2}) * (1 + \frac{1}{2}) =$

$1 + \frac{k}{2} + \frac{1}{2} + \frac{k}{4} >$

$1 + \frac{k}{2} + \frac{1}{2}=$

$1 + \frac{k+1}{2}$

Answer 2

You are trying to establish, from $(1+1/2)^k \ge 1 + k/2$ that $(1 + 1/2)^{k+1} \ge 1 + (k+1)/2$. That is, you are given a statement of the form: $$a \ge b$$ and are trying to establish a statement of the form $$a\cdot c \ge d$$

So you need to establish $b \ge \frac dc$, that is, you need to establish:

$$1 + \frac k2 \ge \dfrac{1 + \dfrac{k + 1}{2}}{1 + \dfrac 12}$$

Should be straightforward.

Answer 3

Continue expanding the product.

$$(1 + \frac{1}{2})^{k+1} =(1 + \frac{1}{2})^k \cdot (1 + \frac{1}{2}) \ge (1+\frac{k}2)(1 + \frac{1}{2}) = 1+\frac{k}2 + \frac12+\frac{k}{4}>1+\frac{k+1}{2}$$