Question

Using geometric method to prove AM GM inequality

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

Steven Zheng Steven Zheng posted 1 week ago


Answer 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}

Steven Zheng Steven Zheng posted 1 week ago


Answer 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}

Steven Zheng Steven Zheng posted 1 week ago


Answer 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.

Steven Zheng Steven Zheng posted 6 days ago

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