If x, y, z are positive numbers in AP then
y2≥xz
y≥2xz−−√
x+y2y−x+y+z2y−z
has the minimum value 2
To determine the minimum value of the expression, we can use calculus. Let's denote the common difference of the AP as d. We have:
x = y - d
z = y + d
Substituting these values into the expressions, we get:
y^2 ≥ (y - d)(y + d)
y ≥ √(y^2 - d^2)
y ≥ √((y - x)(y - z)) (since x = y -d and z = y + d)
Let's denote √((y - x)(y - z)) as a. Now, we can rewrite the expressions:
y ≥ a
x + y^2/(y - x) + y + z^2/(y - z) = (y^3 - xyz + y^3)/(y^2 - xy - xz - yz + xz) = (2y^3 - xyz)/(y^2 - yz)
To find the minimum value of this expression, we take the first derivative with respect to y:
d/dy [(2y^3 - xyz)/(y^2 - yz)] = (2y^4 - 2xyz - 2y^3yz - xy^2)/(y^2 - yz)^2
Setting this derivative equal to zero and solving for y, we get:
2y^4 - 2xyz - 2y^3yz - xy^2 = 0
Simplifying, we have:
2y^4 - 2xy(y^2 + yz) - xy^2 = 0
y^4 - xy(y^2 + yz) - xy^2/2 = 0
Since y is positive, we can divide both sides of the equation by xy:
y^3 - y^2 - yz/2 = 0
y^3 - y^2 = yz/2
y(y^2 - y) = yz/2
Since we know that y ≠ 0:
y^2 - y = z/2
Since y and z are in AP, we can write:
z = y + d
Substituting this into the equation above:
y^2 - y = (y + d)/2
2y^2 - 2y = y + d
2y^2 - 3y - d = 0
This is a quadratic equation in terms of y. We want to find the minimum value, so we set the discriminant equal to zero:
(-3)^2 - 4(2)(-d) = 9 + 8d = 0
d = -9/8
Since d is positive, we discard this solution and conclude that there is no minimum value for the given expression.
In summary, the expression does not have a minimum value.