Visualizing 1 + 1/xDate: 10/10/2003 at 21:25:38 From: Mary Subject: logic How can I show that the sum of a positive number and its reciprocal is at least two? Date: 10/11/2003 at 06:06:27 From: Doctor Luis Subject: Re: logic Hi Mary, Adding a positive number x > 0 and its reciprocal 1/x gives you the function f(x) = x + 1/x If you're familiar with calculus, you can see that solving for the extrema points gives you f'(x) = 1 - 1/x^2 = 0 1 = 1/x^2 x^2 = 1 x = 1 (reject negative root since x > 0) Since f"(x) = 2/x^3 is positive for x>0, we know that f(x) is concave upward. This means that the critical point x=1 gives you a minimum. This minimum value is f(1) = 1 + 1/1 = 2. In the following diagram, I've graphed the two functions y = x + 1/x and y = 2. Even if you are not familiar with calculus maybe you can follow the following chain of reasoning: The square of any nonzero real number is positive. As an inspired guess, pick x-1 as the real number to be squared. Then, (x-1)^2 >= 0 (True for all x. Equality holds only for x=1.) x^2 - 2x + 1 >= 0 x^2 + 1 >= 2x Now, let x > 0, since we are only interested in positive numbers. This means that 1/x > 0 too. So, we can multiply by 1/x without reversing the sign of our inequality: (1/x)*(x^2 + 1) >= (1/x)*(2x) x + 1/x >= 2 This proves that the sum of x > 0 and its reciprocal 1/x adds up to at least 2. I hope this helped! Let us know if you have any more questions. - Doctor Luis, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/