Art of Problem Solving Online School

Week 4: Algebra, Part 2MATHCOUNTS/AMC 8 Advanced

For a number of weeks now, we’ve used the following result:

If the sum of two nonnegative numbers is constant, then the product is maximizedwhen the numbers are equal.

Let’s examine this fact a little more closely, in the context of a very important inequality.

1 An Important Inequality

Our result above can be derived from a more general result known as the Arithmetic Mean-Geometric Mean inequality, or AM-GM to fancy folks. It states the following:

Theorem 1.1. AM-GM Inequality: For all nonnegative real numbers a and b, the arithmeticmean of a and b is greater than or equal to the geometric mean of a and b. That is,

a + b2≥

√

ab.

Furthermore, equality of the two sides occurs if and only if a = b.

Proof: Since a and b are nonnegative, both√

a and√

b are well defined. Consider the expression(√a −√

b)2

. Given that this expression is a square, it must be nonnegative:(√a −√

b)2≥ 0.

We can expand the square and get a − 2√

ab + b. Thus

a − 2√

ab + b ≥ 0.

Adding 2√

ab to both sides, and dividing by 2, gives us

a + b2≥

√

ab,

as desired.

Furthermore, if we trace our steps back to the beginning, we see that the inequality will only be an

equality if it is an equality from the very beginning of our proof—that is, if(√

a −√

b)2= 0. Taking

square roots and rearranging, we have√

a =√

b, or a = b. So, equality happens if and only if a = b.�

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Art of Problem Solving Online School

Week 4: Algebra, Part 2MATHCOUNTS/AMC 8 Advanced

2 Two Important Corollaries

A corollary is a mathematical truth that follows almost immediately from a known theorem.Typically, one can prove a corollary in only a few lines.

Problem 2.1. Using AM-GM, prove these corollaries:

(a) If two nonnegative numbers have a constant sum, then their product is maximal when theyare equal.

(b) If two nonnegative numbers have a constant product, then their sum is minimal when theyare equal.

Solution: (a) Suppose our constant sum is S. By AM-GM, we know that if a + b = S, then

ab ≤(

a + b2

)2

=S2

4,

with equality if and only if a = b. Thus, the product of a and b cannot be larger than S2

4 , and is onlyequal to S2

4 (its maximum possible value) when a = b.

(b) Suppose our constant product is P. By AM-GM, we know that if ab = P, then

a + b ≥ 2√

ab = 2√

P,

with equality if and only if a = b. Thus, the sum of a and b cannot be smaller than 2√

P, and is onlyequal to 2

√P (its minimum possible value) when a = b. �

3 Looking Ahead

When you see a result like AM-GM, you should develop a habit of always asking if the resultgeneralizes. (To generalize means to extend the result so that it covers more possibilities in someway.) For example:

Problem 3.1. One way to think of the Pythagorean Theorem is that it tells you that if c is thelength of a diagonal of a rectangle with sides a and b, then a2 + b2 = c2. What do you think thePythagorean Theorem says in three dimensions, rather than two?

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Art of Problem Solving Online School

Week 4: Algebra, Part 2MATHCOUNTS/AMC 8 Advanced

Solution: If we take a rectangular prism with sides a, b, and c, and space diagonal d, then it turnsout that a2 + b2 + c2 = d2. Can you prove this? �

Problem 3.2. What do you think AM-GM looks like for 3 numbers, a, b, and c? What about nnumbers, a1, a2, . . . , an?

Solution: The arithmetic mean of 3 numbers a, b, c isa + b + c

3, and the geometric mean of a, b, c is

3√abc. (Note that it’s a cube root, not a square root!) So AM-GM for three numbers states that

a + b + c3

≥3√

abc,

with equality if and only if a = b = c.

More generally, with n numbers a1, a2, . . . , an, we have

a1 + a2 + · · · + an

n≥

n√a1a2 · · · an,

with equality if and only if a1 = a2 = · · · = an.

However, proving AM-GM for more than 2 numbers is rather tough! �

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