Proving One Function is Greater Than Another

Date: 11/1/96 at 19:12:28
From: brenchi
Subject: Question

3) prove that if n is a positive integer and x > 0 then:
        
        x^n + 1/(x^n)  >  x^(n-1) + 1/[x^(n-1)]

I would really appreciate your help!

   Billy                       

Date: 11/2/96 at 16:39:1
From: Doctor Anthony
Subject: Re: Question

Prove that if n is a positive integer and x > 0 then:
        
        x^n + 1/(x^n)  >  x^(n-1) + 1/[x^(n-1)]

Let x^n = e^y   and  x^(n-1) = e^((n-1)y/n)

 The lefthand side of the original expression = 2cosh(y)
The righthand side of the original expression = 2cosh[(n-1)y/n]

Since (n-1)y/n < y and since cosh(y) is an increasing function for 
y > 0, it follows that the inequality is true.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 11/2/96 at 19:27:56
From: brenchi
Subject: Re: Question

Dear Doctor Anthony,

Thank you very much for your help! But can you show me another way to 
do it that doesn't involve the use of cosh(y)? In Australia, cosh(y) 
is only taught in University, not in grade 12.

   Billy

Date: 11/3/96 at 8:34:21
From: Doctor Anthony
Subject: Re: Questions

We can start the same way as before:

Let x^n = e^y   and x^(n-1) = e^(ky)  where k = (n-1)/n  < 1

We need to show  e^y + e^(-y) > e^(ky) + e^(-ky)

This will be true if e^y + e^(-y) is an increasing function of y.

To make the notation more convenient, I shall be trying to show that 
f(x) = e^x + e^(-x) is an increasing function of x. i.e. if x>0 then 
as x increases so does f(x).

When x=0 we have f(0) = 2    
Now differentiate f(x):

  f'(x) = e^x - e^(-x)

Now e^x is greater than or equal to 1, while e^(-x) is less than or 
equal to 1.  It follows that f'(x) is zero or positive, so f(x) is an 
increasing function 

Going back to the y variable, e^y + e^(-y) > e^ky + e^(-ky).

This means that the original inequality is true.          
 
-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/