# mathematical induction

• Sep 12th 2007, 09:34 AM
PvtBillPilgrim
mathematical induction
I have to prove this inequality using mathematical induction:

(2)(4)(6).....(2n)
---------------- > 1 + (1/3) + (1/5) + .... + 1/(2n-1).
(1)(3)(5)....(2n-1)

I'm given that for positive reals a1, a2, a3,....,an where n >= 2 that
the cartesian product of (1 + ai) from i = 1 to n is greater than 1 + a1 + a2 + .... + an.

I know how to do the inductions with equal signs, but I can't seem to replace all the terms before (n + 1) on the left hand side with the terms on the right hand side (after assuming the statement is true for all n) since they're not equal.

How do you do this one?

Don't worry about the base case.
Thanks for any help.
• Sep 12th 2007, 10:11 AM
red_dog
Let $n=2$:
$(1+a_1)(1+a_2)=1+a_1+a_2+a_1a_2>1+a_1+a_2$
so the inequality is true.
Suppose the inequality is true for $n$ and prove for $n+1$.
That means $(1+a_1)(1+a_2)\ldots(1+a_n)(1+a_{n+1})>1+a_1+a_2+\ ldots+a_n+a_{n+1}$.
Using the statement for $n$ we have
$(1+a_1)(1+a_2)\ldots(1+a_n)(1+a_{n+1})>(1+a_1+a_2+ \ldots a_n)(1+a_{n+1})=$
$=1+a_1+a_2+\ldots+a_n+a_{n+1}+a_{n+1}(1+a_1+a_2+\l dots a_n)>1+a_1+a_2+\ldots+a_n+a_{n+1}$
Using the transitivity yelds the statement for $n+1$.
• Sep 12th 2007, 01:28 PM
PvtBillPilgrim
I appreciate the help, but how do you prove the first inequality?
• Sep 12th 2007, 01:44 PM
Jhevon
Quote:

Originally Posted by PvtBillPilgrim
I appreciate the help, but how do you prove the first inequality?

what first inequality? you mean for n = 2? we need not "prove" that one, since it is obvious (we have one more positive term on the left side than we do on the right). or was that not what you were talking about?
• Sep 12th 2007, 01:54 PM
red_dog
The first inequality can be written as
$\displaystyle (1+1)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5} \right)\ldots\left(1+\frac{1}{2n-1}\right)>1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1 }{2n-1}$
Apply the general inequality taking $a_1=1,a_2=\frac{1}{3},\ldots,a_n=\frac{1}{2n-1}$
Then we have
$\displaystyle (1+1)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{5} \right)\ldots\left(1+\frac{1}{2n-1}\right)>1+1+\frac{1}{3}+\frac{1}{5}+\ldots+\frac {1}{2n-1}>$
$\displaystyle>1+\frac{1}{3}+\frac{1}{5}+\ldots+\fr ac{1}{2n-1}$
and apply transitivity.
• Sep 12th 2007, 03:29 PM
PvtBillPilgrim
Makes sense.

Thanks a lot.
• Sep 13th 2007, 08:00 AM
PvtBillPilgrim
Just a quick question.

Does this last proof use mathematical induction or is it just a proof? It seems it doesn't prove the n + 1 case.
• Sep 13th 2007, 10:23 AM
red_dog
The last proof is a particular case of the general case which was proved by induction.