Arithmetic Mean - Geometric Mean (AM-GM)

The Arithmetic Mean - Geometric Mean Inequality states that for any $n$ non-negative real values $x_1, x_2, \ldots x_n$, the arithmetic mean of these numbers is greater than or equal to their geometric mean, with equality holding if and only if all the values are equal, i.e.,

$\displaystyle \frac { \displaystyle \sum_{i=1}^n x_i} {n} \geq \sqrt[n] { \prod_{i=1}^n x_i } .$

The proof of this statement for two variables is presented in the post Completing the Square. The proof of the general case is given here.

Here are some consequences of AM-GM:

1. If $a$ is a positive real number, then $a + \frac {1}{a} \geq 2$.

2. If $x$ and $y$ are positive real numbers, then $\frac{x}{y} + \frac{y}{x} \geq 2$.

3. If $a$ is a real number (not necessarily positive), then $a^2 + 1 \geq 2 \cdot \vert a \vert \geq 2 \cdot a$.

Worked Examples

1. Show that $2a^3 + b^3 \geq 3a^2 b$ for $a, b >0$.

Solution: We apply the 3-variable version of AM-GM with $x_1= a^3, x_2 = a^3$ and $x_3 = b^3$ to obtain

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

Then we multiply both sides by 3 to obtain $a^3 + a^3 + b^3 \geq 3a^2 b$.

2. Find all real solutions to $2^x + x^2 = 2 - \frac {1}{2^x}$.

Solution: We have $2 \leq 2^x + \frac {1}{2^x} = 2 - x^2$, so $0 \geq x^2$. This implies $x=0$ is the only possible value. Since $2^0 + 0^2 = 1$ and $2 - \frac {1}{2^0} = 1$, we have verified $x=0$ is the only solution.

3. Find all positive real solutions to

$\begin{aligned} 4x + \frac {18}{y} & = 14 \\ 2y + \frac {9}{z} & = 15 \\ 9z + \frac {16}{x} & = 17\\ \end{aligned}$

Solution: By AM-GM, we have $4x + \frac {16}{x} \geq 16, 2y + \frac {18}{y} \geq 12, 9z + \frac {9}{z} \geq 18$. Summing these three inequalities, we obtain

$4x + \frac {16}{x} + 2y + \frac {18}{y} + 9z + \frac {9}{z} \geq 16 + 12 + 18 = 46 .$

Furthermore, summing the three given equations, we obtain

$4x + \frac {16}{x} + 2y + \frac {18}{y} + 9z + \frac {9}{z} = 46.$

Hence, equality must hold throughout, implying $x=2, y=3$ and $z=1$. By substituting these values into the original equations, we see that $(x, y, z) = (2, 3, 1)$ is indeed a solution.

4. [2-variable Cauchy Schwarz Inequality] Show

$(a^2 + b^2)(c^2 + d^2) \geq (ac+bd)^2 .$

Solution 1: Expanding both sides, we can cancel terms $a^2c^2$ and $b^2d^2$, so we need to show that $a^2 d^2 + b^2c^2 \geq 2acbd$. This follows from the 2-variable AM-GM by setting $x_1 = a^2 d^2$ and $x_2 = b^2 c^2$, to obtain

$a^2d^2 + b^2c^2 \geq 2 \lvert abcd \rvert \geq 2 abcd.$

Solution 2: From Completing the Square's Fermat's Two Square Theorem, we have

$(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2 .$

Since squares are non-negative, the right hand side is greater than or equal to $(ac+bd)^2$.

5. Show that if $a, b$, and $c$ are positive real numbers, then

$a^4 + b^4 + c^4 \geq abc(a+b+c).$

Solution: A direct application of AM-GM doesn’t seem to work. Let's consider how we can get terms on the right hand side through AM-GM. To get $a^2bc$, we will need 'more' of $a$ than of $b$ or $c$ (as in Worked Example 1). This gives a hint to try

$a^4 + a^4 + b^4 + c^4 \geq 4 a^2 b c.$

Similarily, we have

$a^4 +b^4 +b^4 +c^4 \geq 4ab^2 c$

and

$a^4 + b^4 + c^4 + c^4 \geq 4abc^2.$

Adding these 3 inequalities and dividing by 4 yields

$a^4 + b^4 + c^4 \geq a^2bc + ab^2c + abc^2 = abc(a+b+c).$