# Inequalities

• Sep 12th 2012, 05:43 AM
GrigOrig99
Inequalities
Can anyone help me confirm if I have solved this correctly?

Many thanks.

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

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

Step 2: Assume the statement is true for n = k, i.e. assume (1 + ax)k > 1 + akx.
We must now show that the statement is true for n = k + 1,
i.e. (1 + ax)k+1 > 1 + a(k + 1)x = 1 + akx + ax
(1 + ax)k+1 = (1 + ax)(1 + ax)k > (1 + ax)(1 + akx)...((1 + ax)k > 1 + akx...assumed)
If (1 + ax)(1 + akx) > 1 + akx + ax, then (1 + ax)k+1 > 1 + akx + ax
(1 + ax)(1 + akx) > 1 + akx + ax
if (1 + ax)(1 + akx) - (1 + akx + ax) > 0
if akx2 + akx + ax + 1 - 1 - akx - ax > 0
if akx2 > 0
if x2[ak] > 0...true when a & x > 0 & n > 1
(1 + ax)k+1 > 1 + akx + ax
Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all a, x & n > 1, n $\displaystyle \in \mathbb{N}$.
• Sep 12th 2012, 06:48 AM
HallsofIvy
Re: Inequalities
Quote:

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

Many thanks.

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

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

Step 2: Assume the statement is true for n = k, i.e. assume (1 + ax)k > 1 + akx.
We must now show that the statement is true for n = k + 1,
i.e. (1 + ax)k+1 > 1 + a(k + 1)x = 1 + akx + ax
(1 + ax)k+1 = (1 + ax)(1 + ax)k > (1 + ax)(1 + akx)...((1 + ax)k > 1 + akx...assumed)
If (1 + ax)(1 + akx) > 1 + akx + ax,

Why the "if" here?

Quote:

then (1 + ax)k+1 > 1 + akx + ax
(1 + ax)(1 + akx) > 1 + akx + ax
if (1 + ax)(1 + akx) - (1 + akx + ax) > 0
if akx2 + akx + ax + 1 - 1 - akx - ax > 0
if akx2 > 0
if x2[ak] > 0...true when a & x > 0 & n > 1
(1 + ax)k+1 > 1 + akx + ax
Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all a, x & n > 1, n $\displaystyle \in \mathbb{N}$.
You have a whole series of "if"s here. Do you actually show any of those are true?

The simplest way to do this is to actually mutiply $\displaystyle (1+ ax)(1+ akx))$. What do you get when you do that?
• Sep 12th 2012, 06:52 AM
Plato
Re: Inequalities
Quote:

Originally Posted by GrigOrig99
Can anyone help me confirm if I have solved this correctly?
Q.
(1 + ax)n > 1 + anx, for a > 0, x > 0 & n $\displaystyle \in \mathbb{N}$
Attempt: Step 1: For n = 1...
Since 1 + ax > 1 + ax, n is true.

Step 2: Assume the statement is true for n = k, i.e. assume (1 + ax)k > 1 + akx.
We must now show that the statement is true for n = k + 1,

This is also known as Bernoull's inequality.

You tend to over-do it and it is hard to read.
\displaystyle \begin{align*}(1+ax)^{k+1}&=(1+ax)(1+ax)^k \\&\ge (1+ax)(1+akx) \\&=1+akx+ax+a^2kx\\ &\ge 1+akx+akx\\&=1+a(k+1)x\end{align*}
• Sep 12th 2012, 07:49 AM
GrigOrig99
Re: Inequalities
I've mentioned this before, but I guess it bears repeating. The format I'm using to answer these questions is coming from the text book itself; an example of which I will provide in an attachment.

Also, in the previous Inequalities question I posted, (1 +2x)(1 + 2kx) was solved by user 'Prove It' as follows:
= 1 + 2kx + 2x + 2kx2

Comparing his solution with that of (1 + ax)(1 + akx), shouldn't it be solved as:
1 + akx + ax + akx2, rather than 1 + akx + ax + a2kx...?