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Proving One Function is Greater Than Another

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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/
```
Associated Topics:
High School Functions

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