Arithmetic Mean - Geometric Mean (AM-GM)

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The Arithmetic Mean - Geometric Mean Inequality states that for any n n non-negative real values x1,x2,xn 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.,

i=1nxini=1nxin. \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 a is a positive real number, then a+1a2 a + \frac {1}{a} \geq 2.

  2. If x x and y y are positive real numbers, then xy+yx2 \frac{x}{y} + \frac{y}{x} \geq 2.

  3. If a a is a real number (not necessarily positive), then a2+12a2a a^2 + 1 \geq 2 \cdot \vert a \vert \geq 2 \cdot a.

 

Worked Examples

1. Show that 2a3+b33a2b 2a^3 + b^3 \geq 3a^2 b for a,b>0a, b >0.

Solution: We apply the 3-variable version of AM-GM with x1=a3,x2=a3 x_1= a^3, x_2 = a^3 and x3=b3 x_3 = b^3 to obtain

a3+a3+b33a3a3b33=a2b. \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 a3+a3+b33a2b a^3 + a^3 + b^3 \geq 3a^2 b.

 

2. Find all real solutions to 2x+x2=212x 2^x + x^2 = 2 - \frac {1}{2^x}.

Solution: We have 22x+12x=2x2 2 \leq 2^x + \frac {1}{2^x} = 2 - x^2, so 0x2 0 \geq x^2. This implies x=0 x=0 is the only possible value. Since 20+02=1 2^0 + 0^2 = 1 and 2120=1 2 - \frac {1}{2^0} = 1, we have verified x=0 x=0 is the only solution.

 

3. Find all positive real solutions to

4x+18y=142y+9z=159z+16x=17 \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+16x16,2y+18y12,9z+9z18 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+16x+2y+18y+9z+9z16+12+18=46. 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+16x+2y+18y+9z+9z=46. 4x + \frac {16}{x} + 2y + \frac {18}{y} + 9z + \frac {9}{z} = 46.

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

 

4. [2-variable Cauchy Schwarz Inequality] Show

(a2+b2)(c2+d2)(ac+bd)2. (a^2 + b^2)(c^2 + d^2) \geq (ac+bd)^2 .

Solution 1: Expanding both sides, we can cancel terms a2c2 a^2c^2 and b2d2 b^2d^2, so we need to show that a2d2+b2c22acbd a^2 d^2 + b^2c^2 \geq 2acbd. This follows from the 2-variable AM-GM by setting x1=a2d2 x_1 = a^2 d^2 and x2=b2c2 x_2 = b^2 c^2, to obtain

a2d2+b2c22abcd2abcd. 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

(a2+b2)(c2+d2)=(ac+bd)2+(adbc)2. (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 (ac+bd)^2.

 

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

a4+b4+c4abc(a+b+c). 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 a2bc a^2bc, we will need 'more' of a a than of b b or c c (as in Worked Example 1). This gives a hint to try

a4+a4+b4+c44a2bc. a^4 + a^4 + b^4 + c^4 \geq 4 a^2 b c.

Similarily, we have

a4+b4+b4+c44ab2c a^4 +b^4 +b^4 +c^4 \geq 4ab^2 c

and

a4+b4+c4+c44abc2. a^4 + b^4 + c^4 + c^4 \geq 4abc^2.

Adding these 3 inequalities and dividing by 4 yields

a4+b4+c4a2bc+ab2c+abc2=abc(a+b+c). a^4 + b^4 + c^4 \geq a^2bc + ab^2c + abc^2 = abc(a+b+c).

Note by Calvin Lin
5 years, 6 months ago

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

Hasan Kassim - 5 years, 2 months ago

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Thanks, Calvin.

Chew-Seong Cheong - 4 years, 11 months ago

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Good job Test iq for Singapore

Bin bin Channel - 1 year, 6 months ago

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Yong Boon, you quite obviously didn't do this

Danushan Dayaparan - 4 years, 5 months ago

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