# Mathematical Induction to Prove Inequalities

• November 28th 2009, 05:14 PM
discretec
Mathematical Induction to Prove Inequalities
I was wondering if anyone could help me out on this problem. Use mathematical induction to establish the following inequality.
(1+1/2)^n>=1+n/2, for n belonging to N (natural numbers)
n=1
(1+1/2)^1>=1+1/2 true for n=1
Assume n=k
(1+1/2)^k>=1+k/2
I am stuck on how to do the n=k+1. Any help is appreciated.
• November 28th 2009, 06:26 PM
tonio
Quote:

Originally Posted by discretec
I was wondering if anyone could help me out on this problem. Use mathematical induction to establish the following inequality.
(1+1/2)^n>=1+n/2, for n belonging to N (natural numbers)
n=1
(1+1/2)^1>=1+1/2 true for n=1
Assume n=k
(1+1/2)^k>=1+k/2
I am stuck on how to do the n=k+1. Any help is appreciated.

The inductive hypothesis is that $\left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2}$ , so:

$\left(1+\frac{1}{2}\right)^{n+1}=\left(1+\frac{1}{ 2}\right)^n\left(1+\frac{1}{2}\right)\buildrel\tex t{ind. hyp.}\over \geq \left(1+\frac{n}{2}\right)\left(1+\frac{1}{2}\righ t)$ $=1+\frac{n+1}{2}+\frac{n}{4}\geq 1+\frac{n+1}{2}$ . QE.D.

Tonio
• November 28th 2009, 06:46 PM
discretec
Quote:

Originally Posted by tonio
The inductive hypothesis is that $\left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2}$ , so:

$\left(1+\frac{1}{2}\right)^{n+1}=\left(1+\frac{1}{ 2}\right)^n\left(1+\frac{1}{2}\right)\buildrel\tex t{ind. hyp.}\over \geq \left(1+\frac{n}{2}\right)\left(1+\frac{1}{2}\righ t)$ $=1+\frac{n+1}{2}+\frac{n}{4}\geq 1+\frac{n+1}{2}$ . QE.D.

Tonio

Can I ask what you mean by QE.D.?
• November 28th 2009, 06:52 PM
tonio
Quote:

Originally Posted by discretec
Can I ask what you mean by Q.E.D.?

Q.E.D. = Quod erat demonstrandum = what had to be demonstrated.

According to a friend of mine, Q.E.D. = Queer and Ethereal Dork (it's tough to love mathematics...**sigh**)

Tonio
• November 28th 2009, 07:08 PM
discretec
Quote:

Originally Posted by tonio
The inductive hypothesis is that $\left(1+\frac{1}{2}\right)^n\geq 1+\frac{n}{2}$ , so:

$\left(1+\frac{1}{2}\right)^{n+1}=\left(1+\frac{1}{ 2}\right)^n\left(1+\frac{1}{2}\right)\buildrel\tex t{ind. hyp.}\over \geq \left(1+\frac{n}{2}\right)\left(1+\frac{1}{2}\righ t)$ $=1+\frac{n+1}{2}+\frac{n}{4}\geq 1+\frac{n+1}{2}$ . QE.D.

Tonio

Thank you for your help. I am going to stare at this one for awhile. These types of induction problems give me so much trouble. (Thinking) lol
• November 28th 2009, 07:09 PM
discretec
Quote:

Originally Posted by tonio
Q.E.D. = Quod erat demonstrandum = what had to be demonstrated.

According to a friend of mine, Q.E.D. = Queer and Ethereal Dork (it's tough to love mathematics...**sigh**)

Tonio

(Giggle)