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# This note has been used to help create the Arithmetic Mean - Geometric Mean wiki

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

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

Note by Calvin Lin
3 years, 6 months ago

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Thanks, Calvin. · 2 years, 11 months ago

very interesting note, it helped me a lot. Thanks! · 3 years, 1 month ago