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:

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

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

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

## 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

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

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## Comments

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TopNewestvery interesting note, it helped me a lot. Thanks!

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Yong Boon, you quite obviously didn't do this

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