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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Thanks, Calvin.

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Yong Boon, you quite obviously didn't do this

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