Proving One Function is Greater Than AnotherDate: 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/ |
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