# Inequalities

• Aug 28th 2012, 05:44 AM
GrigOrig99
Inequalities
Can anyone help me confirm if I have solved this correctly?
Many thanks.

Q.
3n > n2 for n > 2, n $\in \mathbb{N}$.

Attempt:
Step 1: For n = 2,
32 = 9 & 22 = 4
Since 9 > 4, the statement is true for n = 2.

Step 2: Assume the statement is true for n = k, i.e. assume 3k > k2.
We must now show that the statement is true for n = k + 1,
i.e. 3k+1 > (k + 1)2
3k+1 = 31 $\cdot$3k...(3k > k2​...assumed​)
If it can be shown that 3(k2) > (k + 1)2, then 3k+1 > (k + 1)2
3(k2) > (k + 1)2
if 3k2 - (k + 1)2 > 0
if 3k2 - k2 - 2k - 1 > 0
if 2k2 - 2k - 1 > 0
if 2k2 - 2k + 1 - 1 - 1 > 0...completing the square
if 2k2 - 2k + 1 - 2 > 0
if (2k - 1)(k - 1) - 2 > 0...true for k > 2 = 3k+1 > (k + 1)2
Therefore, the statement is true for n = k + 1, if true for n = k. Thus the statement is true for all n > 2, n $\in \mathbb{N}$.
• Aug 28th 2012, 07:17 AM
Plato
Re: Inequalities
Quote:

Originally Posted by GrigOrig99
Q.[/B] 3n > n2 for n > 2, n $\in \mathbb{N}$.

Attempt:
Step 1: For n = 2,
32 = 9 & 22 = 4
Since 9 > 4, the statement is true for n = 2.
Step 2: Assume the statement is true for n = k, i.e. assume 3k > k2.
We must now show that the statement is true for n = k + 1,

Here is a way.
$3^{K+1}=3(3^K)\ge3(K^2)= K^2+K^2+K^2\ge K^2+2K+1=(K+1)^2$
• Aug 28th 2012, 07:32 AM
GrigOrig99
Re: Inequalities
So that's an alternative? Is the other approach I've shown also correct?
• Aug 28th 2012, 08:30 AM
emakarov
Re: Inequalities
Quote:

Originally Posted by GrigOrig99
3k+1 = 31 $\cdot$3k...assumed

Why assume this since this is immediately provable?

Quote:

Originally Posted by GrigOrig99
3(k2) > (k + 1)2 = if 3k2 - (k + 1)2 > 0

I don't understand this notation where an inequality "= if".

Quote:

Originally Posted by GrigOrig99
if 2k2 - 2k + 1 - 2 > 0 = if (2k - 1)(k - 1) - 2 > 0

2k2 - 2k + 1 ≠ (2k - 1)(k - 1).
• Aug 28th 2012, 10:58 AM
GrigOrig99
Re: Inequalities
Ok, I think I understand now. Thanks guys.
• Aug 28th 2012, 10:58 AM
HallsofIvy
Re: Inequalities
Why did you start with k= 2? Even after you have completed the last part, you would not have proved true for all N. You would not have proved it was true for k= 1.
• Aug 28th 2012, 02:26 PM
GrigOrig99
Re: Inequalities
I'm posting an example from the book to demonstrate my approach. The author uses the values specified in the question to completion, hence my sticking with the 2 throughout my attempt. The book also introduces 'completion of the square' but only uses simplistic examples, which I have now had to read up on independently of the book, in lieu of better examples. This entire chapter has been a complete mess, compared to the others!!
• Aug 29th 2012, 04:52 AM
emakarov
Re: Inequalities
I think the problem in post #1 starts with n = 2 because the proof method of the inductive step shown in a similar problem in post #7's attachment works only for n ≥ 2. The inductive step shows that $3^k > k^2$ implies $3\cdot 3^k > (k + 1)^2$. The conclusion is true for k = 1; however, if we replace $3^k$ by $k^2$ in the conclusion's left-hand side, then the resulting inequality is true only for k ≥ 2. This follows from the fact that the positive root of 2k² - 2k - 1 = 0 is $\frac{1+\sqrt{3}}{2}\approx 1.37$. Of course, it it possible to break the proof of the inductive step into two cases: k = 1 and k ≥ 2.

Speaking about quadratic equations and inequalities, I personally never complete the square or guess the roots based on Vieta's formulas; I always use the quadratic formula.
• Aug 29th 2012, 05:40 AM
Bingk
Re: Inequalities
I think what you've done is okay (although as some pointed out, you need to be more careful with your notation), except for one thing, which is that you did not complete the square correctly. You used the method for when the coefficient of k^2 is 1 ... in this case it is 2. You'd end up with roots, which I'm not sure would be obvious. You could try this also:

From 2k^2 - 2k - 1 > 0
2k^2 - 2k > 1
2k ( k - 1 ) > 1 is true for k > 1