Arithmetic and geometric means inequality
The arithmetic and geometric means inequality, or the AM–GM 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 } \leq \sqrt[ n ]{ 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 △APC and △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 △PCO is a right triangle
PC^2 = PO^2-OC^2
PC^2 = PO^2-r^2
\quad \quad\ = (PO-r)(PO+r)
\quad \quad\ = PA\cdotp PB
\quad \quad\ = ab
\therefore c^2 = ab , c=\sqrt{ab}
On the other hand
\dfrac{a+b}{2} = \dfrac{PA+PB}{2}
\quad\quad\ \ \ =\dfrac{(PO-r)+(PO+r)}{2}
\quad\quad\ \ \ =PO
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 AM–GM inequality of three terms
\dfrac{a+b+c}{3} \geq \sqrt[3]{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