For every two nonnegative real numbers and the following inequality holds:
The inequality above is called the inequality of arithmetic and geometric means.
The equality is valid if and only if .
Let’s assume that the statement is valid. Then we can write it in the following form:
is always greater than or equal to , therefore, the statement is true.
Assume that , without loss generality. Construct a circle with the center at point and diameter , whereby and . It follows that the radius is .
Furthermore, construct the perpendicular bisector . The length of the perpendicular bisector is equal to , what we can show by using the Pythagorean theorem.
The radius is the arithmetic mean and the altitude is the geometric mean.
Since the triangle is the right triangle, then its hypotenuse has the greater length than the leg of the same triangle.
Another geometrical approach
Assume that . Construct the square with the length of the segment and right-angled triangles with legs of the length and (blue triangles). The total area of the blue triangles is . Since the red triangles are reflections across hypotenuses of the blue ones, then the total area of the red triangles is also . Therefore, the total area of all triangles is .
The area of the given square is , therefore, finally we have
In general, for , , is valid:
If , then . Prove it!
After squaring the condition , we obtain:
By using the inequality of arithmetic and geometric means on two numbers, we have:
Analogously, we obtain:
By adding the inequality above, we obtain:
By the condition, we have , that is
Prove that for and is valid:
From the condition, we have:
Now we have:
Now we use the inequality of arithmetic and geometric means for three numbers:
Finally we have:
Prove that the following is valid:
We use the inequality of arithmetic and geometric means:
Prove that for all real numbers is valid:
We will use the inequality of arithmetic and geometric means:
Now we have: