The Art and Craft of Problem Solving

176 CHAPTER 5 ALGEBRA

Now our problem is equivalent to determining the relative order of f( 1998) and f( 1999).

How does this function grow? We have

x 1

f(x) =

x+1

=

1+^1 '

(6)

x

and now it is easy to check that as x > 0 increases, the 1/ x term decreases, causing f{x)

to increase (all we are using is the fundamental principle 5.5.3 that says informally, "If

the denominator increases, the fraction decreases, and vice versa"). In other words,

f{x) is monotonically increasing for positive x.

In any event,f( 1998) < f(1999). •

Notice how we actually made an expression uglier in (6). The right-hand side of

(6) is certainly unpleasant from a typographical standpoint, but it makes it much easier

to analyze the behavior of the function.

Here are a few more for you to tryon your own. Most are fairly easy exercises.

5.5.11 Let al,a 2 ,'" ,an be real numbers. Prove that if '2.ar = 0, then a) = a 2 = ... =

an =0.

5.5.12 The average principle. Let al ,a 2 , ... ,an be real numbers with sum S. Prove

(as rigorously as you can!) that either the ai are all equal, or else at least one of the ai

is strictly greater than the average value S / n and at least one of the ai is strictly less

than the average value.

5.5.13 The notation n!(

k) means take the factorial of n k times. For example, n!( 3

)

means ((n!)!)!. Which is bigger, 1999 !(^2000 ) or 2000!(^1999 l?

101999 + 1 101998 + 1

5.5.14 Which is bigger,

102000 + 1

or

101 999 + 1

?

5.5. 15 Which is bigger, 2oo0! or 10002 oOO?

5.5.16 Which is larger, 19991999 or 20oo^1998?

The AM-GM Inequality

Return to Example 5.5. 10 on page 17 5. The alleged inequality

1998 1999

?

1999 2000

is equivalent (after mUltiplying both sides by (1999· 2000 ) to

1998 · 2000? 19992.

Your intuition probably tells you that the question mark should be replaced with "<,"

for the left-hand side is the area of a rectangle that is not quite a square, while the

right-hand side is the area of a square with the same perimeter as the rectangle (namely

4. 1999). It makes sense that given a fixed amount of fencing, the rectangle of max­

imum area is a square. Indeed, you have probably done this problem as an easy cal­

culus exercise. The underlying principle is very simple mathematically-calculus is

definitely not needed-yet amazingly fruitful.