Applying the Arithmetic Mean Geometric Mean Inequality

We will look at the following 5 general ways of using AM-GM:

1. direct application to an inequality
2. application on each term in a product
3. application on terms obtained after rearranging creatively
4. given an equation, use it to determine extra conditions
5. the inequality case.

The simplest way to apply AM-GM is to apply it immediately on all of the terms. For example, we know that for non-negative values,

$$\frac{x + y } { 2} \geq \sqrt{xy}, \ \ \frac{ x+y+z} { 3} \geq \sqrt[3] { xyz} ,\ \ \frac{w+x+y+z} { 4} \geq \sqrt[4]{wxyz} .$$

Let $a$ and $b$ be positive real numbers. Show that $$\frac {a}{b} + \frac {b}{a} \geq 2.$$

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By setting $x = \frac {a}{b}$ and $y = \frac {b}{a}$ and applying $\frac{x + y } { 2} \geq \sqrt{xy},$ we get

$$\frac {a}{b} + \frac {b}{a} \geq 2 \sqrt{ \frac {a}{b} \cdot \frac {b}{a} } = 2.\ _\square$$

For positive reals $a, b, c$, show that

$$\frac{ a^2 } { bc } + \frac{ b^2 } { ca } + \frac{ c^2 } { ab } \geq 3 .$$

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By setting $x_1 = \frac{ a^2 } { bc}, x_2 = \frac{ b^2}{ca} , x_3 = \frac{ c^2 } { ab}$, we get that

$$\frac{ a^2 } { bc } + \frac{ b^2 } { ca } + \frac{ c^2 } { ab } \geq 3 \sqrt[3] { \frac{ a^2 \times b^2 \times c^2}{bc \times ca \times ab } } = 3. \ _\square$$

Prove that if the product of $n$ positive real numbers is $1$, then their sum is at least $n$.

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Let's call those reals $a_1,a_2,...,a_n$. We are given that $a_1a_2\cdots a_n=1$. Then by AM-GM inequality on $n$ variables we have $$\dfrac{a_1+a_2+\cdots +a_n}{n}\ge \sqrt[n]{a_1a_2\cdots a_n}=1\implies a_1+a_2+\cdots+a_n\ge n,$$ which was what we wanted. $_\square$

If $\left\{b_1,...,b_n\right\}$ is a permutation of the sequence $\left\{a_1,...,a_n\right\}$ of positive reals, then prove that

$$\dfrac{a_1}{b_1}+\dfrac{a_2}{b_2}+\cdots+\dfrac{a_n}{b_n}\ge n.$$

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Notice that $a_1a_2\cdots a_n=b_1b_2\cdots b_n$ since $b$ is a permutation of $a$. Then by AM-GM inequality

$$\dfrac{a_1}{b_1}+\dfrac{a_2}{b_2}+\cdots+\dfrac{a_n}{b_n}\ge n\sqrt[n]{\dfrac{a_1}{b_1}\dfrac{a_2}{b_2}\cdots\dfrac{a_n}{b_n}}=n\sqrt[n]{\dfrac{a_1a_2\cdots a_n}{a_1a_2\cdots a_n}}=n.\ _\square$$

In this approach, we simplify terms of a product first, before multiplying them together.

Show that $( 1 + a^2 ) ( 1 + b^2 ) \geq 4 a b$ for positive reals $a, b$.

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By AM-GM, we have $1 + a^2 \geq 2a$ and $1 + b^2 \geq 2b$. Hence,

$$( 1 + a^2 ) ( 1 + b^2 ) \geq 2a \times 2b = 4ab.\ _\square$$

Note: We can expand the LHS, and then apply AM-GM directly on all the terms

Show that for positive reals $x,y,z$ we have

$$\left(x^2y+y^2z+z^2x\right)\left(xy^2+yz^2+zx^2\right)\ge 9x^2y^2z^2.$$

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Notice that by AM-GM inequality on $3$ variables we have

$$x^2y+y^2z+z^2x\ge 3\sqrt[3]{x^2y\cdot y^2z\cdot z^2x}=3\sqrt[3]{x^3y^3z^3}=3xyz,$$ $$xy^2+yz^2+zx^2\ge 3\sqrt[3]{xy^2\cdot yz^2\cdot zx^2}=3\sqrt[3]{x^3y^3z^3}=3xyz.$$

Multiplying these two, we get

$$\left(x^2y+y^2z+z^2x\right)\left(xy^2+yz^2+zx^2\right)\ge 9x^2y^2z^2$$

as desired. $_\square$

Let $a$ and $b$ be positive real numbers. Show that

$$\frac {1}{ab} \geq \frac {4}{(a+b)^2}.$$

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Cross multiplying both sides, it is equivalent to show that

$$(a+b)\cdot(a+b) = (a+b)^2 \geq 4 ab.$$

To prove this, we apply AM-GM on both of the terms on the LHS:

$$(a+b) \cdot (a+b) \geq 2 \sqrt{ab} \cdot 2 \sqrt{ab} = 4 ab.\ _\square$$

This is the most common way that AM-GM is used, especially in solving Olympiad problems. It can be tricky because it requires you to be innovative and creative in selecting the terms to be used.

Let $a, b, c$ be positive real numbers. Show that

$$a^3 + b^3 + c^3 \geq a^2 b + b^2 c + c^2 a.$$

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We have

$$a^3 + a^3 + b^3 \geq 3 \sqrt[3]{ a^3 a^3 b^3 } = 3 a^ 2 b , \\ b^3 + b^3 + c^3 \geq 3 \sqrt[3] { b^3b^3c^3 } = 3 b^2 c, \\ c^3 + c^3 + a^3 \geq 3 \sqrt[3] { c^3c^3a^3 } = 3 c^2 a .$$

Adding up these inequalities, we get that

$$3 a^3 + 3b^3 + 3 c^3 \geq 3 a^2 b + 3 b^2 c + 3c^2 a .$$

Hence, the result follows when we divide throughout by 3. $_ \square$

This approach generalizes to give us the Muirhead inequality.

Let $a, b, c$ be positive real numbers. Show that

$$\frac{ a^2 } { b^2 } + \frac{ b^2 } { c^2 } + \frac{ c^2 } { a^2 } \geq \frac { b}{a} + \frac{ c}{b} + \frac{ a}{c} .$$

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Applying AM-GM on $\frac{a^2}{b^2 }$ and $\frac{ b^2 } { c^2 }$, we obtain

$$\frac{ a^2 }{ b^2 } + \frac{ b^2 } { c^2 } \geq 2 \sqrt{ \frac{ a^2 b^2 } { b^2 c^ 2} } = 2 \frac{ a}{c}.$$

Observe that this last term is one of the terms on the RHS of our inequality, which is a great news for us. Similarly, we have

$$\begin{aligned} \frac{ b^2 } { c^2 } + \frac{ c^2 } {a^2 } &\geq 2 \frac { b}{a},\\ \frac{ c^2 } { a^2 } + \frac{ a^2 } {b^2 } &\geq 2 \frac { c}{b}. \end{aligned}$$

Adding up these 3 inequalities and dividing by 2, we get that

$$\frac{ a^2 } { b^2 } + \frac{ b^2 } { c^2 } + \frac{ c^2 } { a^2 } \geq \frac { b}{a} + \frac{ c}{b} + \frac{ a}{c} . \ _\square$$

The above example shows us the power of being creative in the selection. If we tried applying AM-GM directly on all of the terms, we would merely get

$$\frac{ a^2 } { b^2 } + \frac{ b^2 } { c^2 } + \frac{ c^2 } { a^2 } \geq 3 \sqrt[3] { \frac { a^2b^2c^2}{b^2c^2a^2}} = 3.$$

However, since

$$\frac { b}{a} + \frac{ c}{b} + \frac{ a}{c} \geq 3 \sqrt[3] { \frac{ bca } { abc } } = 3,$$

we cannot conclude that the inequality is true.

If $a,b,c$ are positive real numbers, then prove that

$$\dfrac{c}{a}+\dfrac{a}{b+c}+\dfrac{b}{c} \geq 2.$$

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We clearly see that direct application of AM-GM won't help. So, we need to transform it a little. Let's try to add $1$ on both sides, and try to show

$$\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}+1 \geq 3 .$$

This follows immediately from AM-GM, since

$$\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c} \geq 3 \sqrt[3] { \frac{ c}{a} \times \frac{a}{b+c} \times \frac{b+c} { c} } = 3.\ _\square$$

Let $a > b$ be positive real numbers. What is the minimum value of

$$a + \frac {1} {b(a-b)}?$$

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We rewrite the expression as $$b + (a-b) + \frac {1}{b(a-b)}.$$ Applying AM-GM shows that $$b + (a-b) + \frac {1}{b(a-b)} \geq 3 \sqrt[3]{ b\times (a-b) \times \frac {1}{b(a-b)}}=3.$$ Equality is achieved when $b=a-b= \frac {1}{b(a-b)}$, or when $a=2$ and $b=1$. Hence, the minimum of the expression is 3. $_\square$

For $x,y,z\in\mathbb{R}^+$ prove that $x^2+y^2+z^2\ge x\sqrt{y^2+z^2}+y\sqrt{x^2+z^2}$.

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By AM-GM inequality $$x^2+\left(y^2+z^2\right)\ge 2\sqrt{x^2\left(y^2+z^2\right)}=2x\sqrt{y^2+z^2},$$ $$y^2+\left(x^2+z^2\right)\ge 2\sqrt{y^2\left(x^2+z^2\right)}=2y\sqrt{x^2+z^2}.$$ Adding these two, we get $2\left(x^2+y^2+z^2\right)\ge 2x\sqrt{y^2+z^2}+2y\sqrt{x^2+z^2}$, dividing which by $2$ gives the desired result. $_\square$

For all positive real values $a,b,c, d$, prove that

$$\frac{a^2}{b} +\frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a} \geq a+b+c+d.$$

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Let's consider that $$\begin{aligned} \frac{a^2}{b} +\frac{a^2}{b} +\frac{b^2}{c}+c  &\geq 4\sqrt[4]{\frac{a^2}{b} \frac{a^2}{b} \frac{b^2}{c} c} \\ &= 4a\\ 2\frac{a^2}{b} + \frac{b^2}{c} +c &\geq 4a. \end{aligned}$$ Do it for the others to obtain $$\begin{aligned} 2\frac{b^2}{c} + \frac{c^2}{d} +d &\geq 4b\\ 2\frac{c^2}{d} + \frac{d^2}{a} +a &\geq 4c\\ 2\frac{d^2}{a} + \frac{a^2}{b} +b &\geq 4d. \end{aligned}$$ Adding all of them, we will get $$\begin{aligned} 3\left(\frac{a^2}{b} + \frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a}\right) +(a+b+c+d) &\geq 4(a+b+c+d)\\ 3\left(\frac{a^2}{b} +\frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a}\right) &\geq 3(a+b+c+d)\\ \frac{a^2}{b} +\frac{b^2}{c} +\frac{c^2}{d} +\frac{d^2}{a} &\geq a+b+c+d. \ _\square \end{aligned}$$

It could initially seem strange that we can use an inequality, to give us information about an equation. In this approach, we show that the equation forces us to have equality, which then allows us to conclude that we are in the equality case.

This approach is very similar to applying the trivial inequality. For example, if we know that $a$ and $b$ are real numbers such that $( a-b) ^ 2 = 0$, then we can immediately conclude that $a=b$. This is because the trivial inequality states that $x^2 \geq 0$, with equality only when $x = 0$.

Find all pairs of positive real numbers such that $a^3 + b^3 + 1 = 3 ab$.

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Applying AM-GM to $a^3 + b^3 + 1$, we get that

$$a^3 + b^3 + 1 \geq 3 \sqrt[3]{ a^3 \times b^3 \times 1 } = 3 ab .$$

Hence, we are in the equality case, and can conclude that $a = b = 1$. $_\square$

To understand the power of this approach, try finding another solution to the above example.

(Britain '96) Find all positive real solutions $(a, b, c, d)$ to the following system of equations:

$$a + b + c + d = 12, \ \ abcd = 27 + ab + ac + ad + bc + bd + cd.$$

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Hint 1: Show that $\sqrt{abcd} \leq 9$.
Hint 2: Show that $\sqrt{abcd} \geq 9$.

Proof of hint 1: Applying AM-GM, $3 = \frac { a + b + c + d}{4} \geq \sqrt[4]{abcd}$. Since both sides are positive, we may square both sides to get $9 \geq \sqrt{abcd}$.

Proof of hint 2: Notice that hint 1 only uses the first equation, so we would have to use the second equation in this. Applying AM-GM, we get that $ab+ac+ad+bc+bd+cd \geq 6 \sqrt{abcd}$. Hence, $\left(\sqrt{abcd}\right) ^2 -6 \sqrt{abcd} - 27 \geq 0$, or that $\left(\sqrt{abcd} - 9\right)\left(\sqrt{abcd}+3\right) \geq 0$. Since the second term of the product is always positive, this implies $\sqrt{abcd} -9 \geq 0$. $_\square$

From hints 1 and 2, $\sqrt{abcd}=9$. Hence, equality must hold in the AM-GM used in hint 1 (or hint 2), which implies that $a = b= c =d$. Since $a + b + c + d = 12$, this gives $a =b=c=d=3$ as the only possible solution. A quick check shows that this is indeed a solution. $_\square$

To show that a strict inequality holds, we simply show that the equality case does not apply in this scenario.

For integers $n > 1$, show that

$$\left( \frac{1}{2} + \frac{ 2}{3} + \cdots + \frac{ n } { n+1 } \right) ^ n > \frac{ n^n } { n+1 }.$$

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Applying AM-GM, we get that $$\frac{1}{2} + \frac{ 2}{3} + \cdots + \frac{ n } { n+1 } \geq n \sqrt[n] { \frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac { n}{n+1} } = n \sqrt[ n ] { \frac{1}{n+1} } .$$ Raising both sides to the $n^\text{th}$ power, we obtain $$\left( \frac{1}{2} + \frac{ 2}{3} + \cdots + \frac{ n } { n+1 } \right) ^ n \geq \frac{ n^n } { n+1 }.$$ It is obvious that the equality case does not apply since $\frac{1}{2} \neq \frac{2}{3}$. $_\square$

Let $n>1$ be an integer. Show that

$$n! < \left( \frac {n+1}{2} \right)^n.$$

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Applying AM-GM to the terms $x_1=1, x_2=2, \ldots, x_n=n$, we have

$$\begin{array}{rl } \frac {1+2+\cdots + n}{n} & \geq \sqrt[n]{1 \times 2 \times \cdots \times n} \\ \dfrac { \frac {n(n+1)}{2}}{n} & \geq \sqrt[n]{ n!} \\ \left( \frac {n+1}{2} \right)^n & \geq n!. \end{array}$$ Since $n>1$, the terms $x_i$ are not all equal, and hence equality does not hold. As such, this shows that $$\left( \frac {n+1}{2} \right)^n > n!.\ _\square$$

Factorization followed by application of AM-GM inequality:

What is the minimum value that the expression $$\dfrac{x^2y^2 - 2 x^2y +2x^2 +2xy-2x+1}{x^2y + x},$$ where $x,y \in \mathbb{R^+}$, can attain?

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We can rewrite the expression as $$\begin{aligned} \dfrac{ x^{2}y^{2} + 2xy + 1 - 2x^{2}y + 2x^{2} - 2x}{x(xy + 1)} &= \dfrac{ (xy + 1)^{2} - 2x(xy + 1 - x)}{x(xy + 1)}\\ &= \dfrac{xy + 1}{x} + \dfrac{2x}{xy + 1} - 2. \qquad \qquad (1) \end{aligned}$$ By the AM-GM inequality $$\frac{\frac{xy + 1}{x} + \frac{2x}{xy + 1}}{2} \geq \sqrt{2},$$ so the minimum value of $(1)$ is $2\sqrt{2} - 2 \approx 0.828.\ _\square$

Problem credit - Brilliant.org

Here are some questions to try. Think about which of the 5 ways you can use to apply to a particular question.

1) Let $a, b$ and $c$ be positive real numbers. Show that

$$(a+b+c)(ab+bc+ca) \geq 9 abc .$$

2) Let $a$ and $b$ be positive real numbers. Show that

$$\frac {1}{a} + \frac {1}{b} \geq \frac {4}{a+b}.$$

3) Given $n$ positive real numbers $x_1, x_2, \ldots, x_n$, show that

$$\frac {x_1 ^2}{x_2} + \frac {x_2^2}{x_3} + \cdots + \frac {x_n^2}{x_1} \geq x_1 + x_2 + \cdots + x_n .$$

4) (Nesbitt's Inequality) Given 3 positive reals $a, b$ and $c$, show that

$$\frac {a}{b+c} + \frac {b}{a+c} + \frac {c}{a+b} \geq \frac {3}{2} .$$

5) Given a real value $x$ such that $0 \leq x \leq 1$, show that

$$\frac {x^2}{x+1} + \frac {(1-x)^2}{2-x} \geq \frac {1}{3} .$$

6) Let $x$ and $y$ be positive real numbers satisfying $x^3 +y^3 = x-y$. Show that

$$x^2 + 4y^2 < 1.$$

7) Let $n$ be a positive integer. Show that $\left(1 + \frac {1}{n+1}\right)^{n+1} > \left(1+\frac {1}{n}\right)^{n}.$ (Hint: How do we apply AM-GM? Which side is the GM? What is the value of $n$ in the AM-GM? Can we create a set of $n$ terms whose product is the GM?)

8) (*) Given 3 positive reals $a, b$ and $c$ such that $a+b+c=1$, show that $a^a b^b c^c+ a^b b^c c^a + a^c b^a c^b \leq 1$.

9) Let $a,b,c$ be positive real numbers such that $abc=1.$ Prove that

$$\dfrac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{b+a}{\sqrt{c}}≥\sqrt{a}+\sqrt{b}+\sqrt{c}+3.$$