Visualizing 1 + 1/x

Date: 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/