’╗┐ Arithmetic and geometric means inequality

# Arithmetic and geometric means inequality

amgm
inequality

The arithmetic and geometric means inequality, or the AMŌĆōGM inequality, states that the arithmetic mean of a list of positive real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is equal to each other.

\dfrac{x_1+x_2+\dots+x_n}{n} \geq \sqrt[n]{x_1\cdotp x_2 \dots x_n }

and that equality holds if and only if x_1 = x_2 = \dots = x_n

or

### Derivation 1

Draw an inscribed right triangle within a half circle with its hypotenuse coinciding with the diameter of the circle AB .

C is any point on the circumference of the circle. Construct a perpendicular from C and intersect with AB at P.

As Ō¢│APC and Ō¢│CPB are similar triangles

\therefore \dfrac{AP}{CP}= \dfrac{CP}{BP} , CP = \sqrt{AP\cdotp BP }

On the other hand, AP + BP = AB

\therefore \dfrac{ AP + BP}{2} = OQ

As OQ is the radius of the circle, it is always larger than or equal to CP

\therefore \dfrac{ AP + BP}{2} \geq \sqrt{AP\cdotp BP }

or

\dfrac{a+b}{2} \geq \sqrt{ab}

### Derivation 2

Plot 4 rectangles of the same size inside the square with their sides equal to a and b respectively.

The area of the square is always larger than or equal to the total size of the 4 rectangles. Therefore,

(a+b)^2\geq 4ab

\therefore \dfrac{a+b}{2} \geq \sqrt{ab}

### Derivation 3

This method uses the power of a circle to prove AM-GM inequality.

Construct a tangent line from point P to the circle at a point . Draw a line connecting point P and the center of the circle O , which intersects with the circle at point A and B .

Let a = PA , b = PB , c = PC ,

we get c^2 = ab

How to prove the conclusion?

Well, as Ō¢│PCO is a right triangle

PC^2 = PO^2-OC^2

PC^2 = PO^2-r^2

\therefore c^2 = ab , c=\sqrt{ab}

On the other hand

\dfrac{a+b}{2} = \dfrac{PA+PB}{2}

As PO is the hypotenuse of the triangle PCO

PO is always larger than PC

\therefore \dfrac{a+b}{2} >\sqrt{ab}

When r \to 0 , PA= PB, the 2 sides of the inequality tends to be equal.

### Derivation 4

To prove the AMŌĆōGM inequality of three terms

\dfrac{a+b+c}{3} \geq \sqrt{abc}

Let a = x^3, b =y^3, c=z^3

The inequality is converted to

x^3+y^3+z^3\geq 3xyz

Using the identity

x^3+y^3+z^3 - 3xyz

= (x+y+z)(x^2+y^2+z^2-xy-yz-xz)

=\dfrac{1}{2} (x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2xz)

=\dfrac{1}{2} (x+y+z)[(x-y)^2+(y-z)^2+(x-z)^2]

\geq 0

\therefore x^3+y^3+z^3\geq 3xyz

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