AlgebraDifficulty 25-80686 tagged problems

AM-GM

AM-GM is the habit of using the arithmetic-geometric mean inequality to bound positive expressions and locate equality. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\frac{x_1 + x_2 + \cdots + x_n}{n} \;\ge\; \sqrt[n]{x_1 x_2 \cdots x_n}$$

For nonnegative numbers, the arithmetic mean is always at least the geometric mean, with equality exactly when all the numbers are equal.

Why It Works

1
Take the two-variable case $\frac{a + b}{2} \ge \sqrt{ab}$ for $a, b \ge 0$.
2
Since a square is never negative, $\left(\sqrt{a} - \sqrt{b}\right)^2 \ge 0$.
3
Expanding gives $a - 2\sqrt{ab} + b \ge 0$, i.e. $a + b \ge 2\sqrt{ab}$.
4
Dividing by $2$ yields $\frac{a+b}{2} \ge \sqrt{ab}$, with equality iff $a = b$. The same idea generalizes to $n$ terms.

Worked Examples

Example 1

For $x > 0$, find the minimum value of $x + \dfrac{9}{x}$.

1
The two terms $x$ and $\frac{9}{x}$ are positive, so apply AM-GM to them.
2
$x + \frac{9}{x} \ge 2\sqrt{x \cdot \frac{9}{x}} = 2\sqrt{9} = 6$.
3
Equality holds when $x = \frac{9}{x}$, i.e. $x^2 = 9$, so $x = 3$.

Answer:

Minimum value $6$, attained at $x = 3$.

Warm-up

Two positive reals satisfy $x + y = 10$. What is the largest possible value of the product $xy$?

1
By AM-GM, $\dfrac{x + y}{2} \ge \sqrt{xy}$, so $\sqrt{xy} \le \dfrac{10}{2} = 5$.
2
Squaring (both sides nonnegative) gives $xy \le 25$.
3
Equality in AM-GM needs $x = y$; with $x + y = 10$ that is $x = y = 5$, where $xy = 25$.

Answer:

Maximum product $25$, at $x = y = 5$.

Contest level

Prove that for all positive reals $a, b, c$, $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge 3$.

1
The three terms $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are all positive, so apply $3$-variable AM-GM.
2
$\dfrac{1}{3}\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right) \ge \sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}} = \sqrt[3]{1} = 1$.
3
Multiply by $3$: the sum is $\ge 3$. Equality requires $\frac{a}{b} = \frac{b}{c} = \frac{c}{a}$, which forces $a = b = c$.

Answer:

$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge 3$, equality iff $a = b = c$.

Olympiad / Challenge

For positive reals with $x + y = 1$, maximize $x^2 y$.

1
Split $x$ into two equal halves so all factors in the product appear in the AM-GM sum: apply AM-GM to the three numbers $\frac{x}{2}, \frac{x}{2}, y$.
2
$\dfrac{\frac{x}{2} + \frac{x}{2} + y}{3} \ge \sqrt[3]{\dfrac{x}{2}\cdot\dfrac{x}{2}\cdot y} = \sqrt[3]{\dfrac{x^2 y}{4}}$. The left side is $\frac{x + y}{3} = \frac{1}{3}$.
3
Cube: $\dfrac{1}{27} \ge \dfrac{x^2 y}{4}$, so $x^2 y \le \dfrac{4}{27}$. Equality needs $\frac{x}{2} = y$, i.e. $x = 2y$; with $x + y = 1$ that gives $x = \frac{2}{3}, y = \frac{1}{3}$, and $x^2 y = \frac{4}{9}\cdot\frac{1}{3} = \frac{4}{27}$.

Answer:

Maximum $\dfrac{4}{27}$, at $x = \tfrac{2}{3},\ y = \tfrac{1}{3}$.

Going Deeper

Generalization: weighted AM-GM says $w_1 x_1 + \cdots + w_n x_n \ge x_1^{w_1}\cdots x_n^{w_n}$ for weights $w_i \ge 0$ summing to $1$; it is itself a special case of Jensen's inequality applied to the concave $\log$.
Where it appears: AMC/AIME optimization ('minimize $x + \frac{k}{x}$'), and as the workhorse first step in olympiad inequalities — often after a clever split (as above) to make the product telescope to a constant.
Pitfall: AM-GM requires NONNEGATIVE terms — applying it to possibly-negative quantities is invalid. Also, a bound is only the true min/max if the equality case is ATTAINABLE under the constraints; if forcing all terms equal violates the constraint, the AM-GM bound is not tight and you must split the terms differently.

Spot the Signal

Look for problems where the key step is using the arithmetic-geometric mean inequality to bound positive expressions and locate equality.
You can describe the hard part as using the arithmetic-geometric mean inequality to bound positive expressions and locate equality, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for am-gm, then rewrite the givens around it.
Name the relation that makes AM-GM legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use AM-GM just because the surface looks familiar; verify the required condition first.
Applying AM-GM because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let am-gm reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using the arithmetic-geometric mean inequality to bound positive expressions and locate equality.

Back to techniques

AlgebraDifficulty 25-80686 tagged problems

AM-GM

Practice this technique Match my level

The Key Result

$$\frac{x_1 + x_2 + \cdots + x_n}{n} \;\ge\; \sqrt[n]{x_1 x_2 \cdots x_n}$$

For nonnegative numbers, the arithmetic mean is always at least the geometric mean, with equality exactly when all the numbers are equal.

Why It Works

1
Take the two-variable case $\frac{a + b}{2} \ge \sqrt{ab}$ for $a, b \ge 0$.
2
Since a square is never negative, $\left(\sqrt{a} - \sqrt{b}\right)^2 \ge 0$.
3
Expanding gives $a - 2\sqrt{ab} + b \ge 0$, i.e. $a + b \ge 2\sqrt{ab}$.
4
Dividing by $2$ yields $\frac{a+b}{2} \ge \sqrt{ab}$, with equality iff $a = b$. The same idea generalizes to $n$ terms.

Worked Examples

Example 1

For $x > 0$, find the minimum value of $x + \dfrac{9}{x}$.

1
The two terms $x$ and $\frac{9}{x}$ are positive, so apply AM-GM to them.
2
$x + \frac{9}{x} \ge 2\sqrt{x \cdot \frac{9}{x}} = 2\sqrt{9} = 6$.
3
Equality holds when $x = \frac{9}{x}$, i.e. $x^2 = 9$, so $x = 3$.

Answer:

Minimum value $6$, attained at $x = 3$.

Warm-up

Two positive reals satisfy $x + y = 10$. What is the largest possible value of the product $xy$?

1
By AM-GM, $\dfrac{x + y}{2} \ge \sqrt{xy}$, so $\sqrt{xy} \le \dfrac{10}{2} = 5$.
2
Squaring (both sides nonnegative) gives $xy \le 25$.
3
Equality in AM-GM needs $x = y$; with $x + y = 10$ that is $x = y = 5$, where $xy = 25$.

Answer:

Maximum product $25$, at $x = y = 5$.

Contest level

Prove that for all positive reals $a, b, c$, $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge 3$.

1
The three terms $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}$ are all positive, so apply $3$-variable AM-GM.
2
$\dfrac{1}{3}\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right) \ge \sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}} = \sqrt[3]{1} = 1$.
3
Multiply by $3$: the sum is $\ge 3$. Equality requires $\frac{a}{b} = \frac{b}{c} = \frac{c}{a}$, which forces $a = b = c$.

Answer:

$\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} \ge 3$, equality iff $a = b = c$.

Olympiad / Challenge

For positive reals with $x + y = 1$, maximize $x^2 y$.

1
Split $x$ into two equal halves so all factors in the product appear in the AM-GM sum: apply AM-GM to the three numbers $\frac{x}{2}, \frac{x}{2}, y$.
2
$\dfrac{\frac{x}{2} + \frac{x}{2} + y}{3} \ge \sqrt[3]{\dfrac{x}{2}\cdot\dfrac{x}{2}\cdot y} = \sqrt[3]{\dfrac{x^2 y}{4}}$. The left side is $\frac{x + y}{3} = \frac{1}{3}$.
3
Cube: $\dfrac{1}{27} \ge \dfrac{x^2 y}{4}$, so $x^2 y \le \dfrac{4}{27}$. Equality needs $\frac{x}{2} = y$, i.e. $x = 2y$; with $x + y = 1$ that gives $x = \frac{2}{3}, y = \frac{1}{3}$, and $x^2 y = \frac{4}{9}\cdot\frac{1}{3} = \frac{4}{27}$.

Answer:

Maximum $\dfrac{4}{27}$, at $x = \tfrac{2}{3},\ y = \tfrac{1}{3}$.

Going Deeper

Generalization: weighted AM-GM says $w_1 x_1 + \cdots + w_n x_n \ge x_1^{w_1}\cdots x_n^{w_n}$ for weights $w_i \ge 0$ summing to $1$; it is itself a special case of Jensen's inequality applied to the concave $\log$.
Where it appears: AMC/AIME optimization ('minimize $x + \frac{k}{x}$'), and as the workhorse first step in olympiad inequalities — often after a clever split (as above) to make the product telescope to a constant.
Pitfall: AM-GM requires NONNEGATIVE terms — applying it to possibly-negative quantities is invalid. Also, a bound is only the true min/max if the equality case is ATTAINABLE under the constraints; if forcing all terms equal violates the constraint, the AM-GM bound is not tight and you must split the terms differently.

Spot the Signal

Look for problems where the key step is using the arithmetic-geometric mean inequality to bound positive expressions and locate equality.
You can describe the hard part as using the arithmetic-geometric mean inequality to bound positive expressions and locate equality, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for am-gm, then rewrite the givens around it.
Name the relation that makes AM-GM legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use AM-GM just because the surface looks familiar; verify the required condition first.
Applying AM-GM because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let am-gm reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using the arithmetic-geometric mean inequality to bound positive expressions and locate equality.