# Inequalities

• Sep 10th 2012, 06:58 AM
GrigOrig99
Inequalities
Can anyone help me confirm if I have solved this correctly?

Many thanks.

Q.
(1 + 2x)n > 1 + 2nx, for x > 0 & n $\displaystyle \in \mathbb{N}$

Attempt: Step 1: For n = 1...
(1 + 2x)1 = 1 + 2x '&' 1 + 2(1)x = 1 + 2x
Since 1 + 2x > 1 + 2x, n is true.

Step 2:
Assume the statement is true for n = k, i.e. assume (1 + 2x)k > 1 + 2kx,
We must now show that the statement is true for n = k + 1,
i.e. (1 + 2x)k+1 > 1 + 2(k + 1)x = 1 + 2kx + 2x
(1 + 2x)k+1 = (1 + 2x)1(1 + 2x)k > (1 + 2x)(1 + 2kx)...((1 + 2x)k > 1 + 2kx...assumed)
If (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x then (1 + 2x)k+1 > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x) > 0
if 2kx2 + 2kx + 2x + 1 - 1 - 2kx - 2x > 0
if 2kx2 > 0
if x2[2k] > 0...true when x > 0 & n > 1
(1 + 2x)k+1 > 1 + 2kx + 2x
Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n > 1, n $\displaystyle \in \mathbb{N}$
• Sep 10th 2012, 07:03 AM
Prove It
Re: Inequalities
Quote:

Originally Posted by GrigOrig99
Can anyone help me confirm if I have solved this correctly?

Many thanks.

Q.
(1 + 2x)n > 1 + 2nx, for x > 0 & n $\displaystyle \in \mathbb{N}$

Attempt: Step 1: For n = 1...
(1 + 2x)1 = 1 + 2x '&' 1 + 2(1)x = 1 + 2x
Since 1 + 2x > 1 + 2x, n is true.

Step 2:
Assume the statement is true for n = k, i.e. assume (1 + 2x)k > 1 + 2kx,
We must now show that the statement is true for n = k + 1,
i.e. (1 + 2x)k+1 > 1 + 2(k + 1)x = 1 + 2kx + 2x
(1 + 2x)k+1 = (1 + 2x)1(1 + 2x)k > (1 + 2x)(1 + 2kx)...((1 + 2x)k > 1 + 2kx...assumed)
If (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x then (1 + 2x)k+1 > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x) > 0
if 2kx2 + 2kx + 2x + 1 - 1 - 2kx - 2x > 0
if 2kx2 > 0
if x2[2k] > 0...true when x > 0 & n > 1
(1 + 2x)k+1 > 1 + 2kx + 2x
Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n > 1, n $\displaystyle \in \mathbb{N}$

I'd do it like this...

\displaystyle \displaystyle \begin{align*} (1 + 2x)^{k + 1} &= (1 + 2x)(1 + 2x)^k \\ &\geq (1 + 2x)(1 + 2kx) \\ &= 1 + 2kx + 2x + 2kx^2 \\ &\geq 1 + 2kx + 2x \end{align*}
• Sep 10th 2012, 08:20 AM
GrigOrig99
Re: Inequalities
Thank you.