Overview

AM-GM is the fastest way to bound expressions with fixed products or sums. It also provides clean equality conditions. In general, there are usually two types of inequality problems that appear.

• The Olympiad version, which is generally the more common version, asks you to prove or find the minimum or maximum.
• The Computational version is far more unique. Usually, they give you an equation that is the equality case, and you have to determine how to use it to your advantage.

Key Ideas

• For nonnegative $a_1, a_2, \ldots, a_n$, $\frac{a_1 + a_2 + \ldots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \ldots a_n}$.
• Equality holds when all variables are equal; that is, when $a_1 = a_2 = \ldots = a_n$, then $\frac{a_1 + a_2 + \ldots + a_n}{n} = \sqrt[n]{a_1 a_2 \ldots a_n}$.

Core Skills

Check Nonnegativity

Firstly, it is important to ensure that one carefully checks that the terms on which the inequality is applied are nonnegative terms. If the variables can be negative, you should try to play around with them a bit more. For instance, reframe the expression with squares before applying it.

Normalize Expressions

Recall, the right-hand side is a product. Rewrite the expression to look like a sum of terms with a fixed product. Then apply AM-GM directly.

Use Equality Cases

The equality condition tells you where the minimum or maximum occurs. Use it to solve for parameters. For instance, in many computational competitions, they won't ask you to prove an inequality. Instead, they may give you the equality, and you have to spot it!

Examples

Example 1. (Basic)

Problem. Compute the minimum possible value of $x + \frac{4}{x}$ for $x>0$.

Solution. By AM-GM, it follows that

So, $x + \frac{4}{x} \ge 4$, with equality at $x=2$. Note that we can derive $x=2$ as the equality occurs when $x = \tfrac{4}{x}$.

Example 2. (More variables)

Problem. Let $a,b,c \ge 0$ be positive real numbers with $abc=1$, compute the minimum of $a+b+c$.

Solution. By AM-GM, $a+b+c \ge 3\sqrt[3]{abc} = 3$.

This occurs when $a = b = c = 1$.

Example 3. (Opposite)

Problem. Let $x$ and $y$ be positive real numbers such that $x+y=10$. Compute the maximum possible value of $xy$.

Solution. By AM-GM,

so $xy \le \boxed{25}$ at $x=y=5$.

Example 4. (Splitting)

Problem. Let $x$ be a positive real number. Compute the minimum possible value of $x^2 + \frac{16}{x}$.

Solution. By AM-GM,

Notably, this occurs when $x^2 = \frac{8}{x}$, and so when $x = 2$.

Example 5. (Unique)

Problem. Let $x,y>2$ be real numbers satisfying $(x - 2)(y - 2) = 81$. Compute the minimum possible value of $x+y$.

Solution. From the given constraints, $x, y > 2$, rewrite this into $x - 2, y - 2, > 0$ and then apply AM-GM:

which becomes $x+y \geq 22$. The equality occurs when $x - 2 = y-2$, which is when $x = y = 11$.

Strategy Checklist

• Confirm all terms are nonnegative.
• Rewrite to expose a fixed product or fixed sum.
• Apply AM-GM and note when equality holds.
• Verify the equality point satisfies the domain.

Common Pitfalls

• Applying AM-GM to negative terms.
• Forgetting to check the equality condition.
• Mixing up minimum and maximum in fixed-sum/product problems.

Practice Problems