Let x,y,z be non-negative real numbers satisfying the condition x+y+z=1. The maximum possible value of
x^3y^3+y^3z^3+z^3x^3
has the form ab where a and b are positive, coprime integers. What is the value of a+b?
To find the maximum possible value of x^3y^3 + y^3z^3 + z^3x^3, we can use the AM-GM inequality.
According to the AM-GM inequality, the arithmetic mean of a set of non-negative numbers is always greater than or equal to the geometric mean.
Let's apply this inequality to the numbers x^3y^3, y^3z^3, and z^3x^3:
(x^3y^3 + y^3z^3 + z^3x^3)/3 ≥ ∛(x^3y^3 * y^3z^3 * z^3x^3)
Simplifying the right side:
(x^3y^3 + y^3z^3 + z^3x^3)/3 ≥ ∛(x^6y^6z^6)
Now we need to find the maximum value of x^6y^6z^6 for the given condition x + y + z = 1.
Using AM-GM on x, y, and z:
(x + y + z)/3 ≥ ∛(xyz)
Since x + y + z = 1, we have:
1/3 ≥ ∛(xyz)
Cubing both sides:
1/27 ≥ xyz
Now, we'll substitute this value into our inequality for x^6y^6z^6:
(x^3y^3 + y^3z^3 + z^3x^3)/3 ≥ ∛(x^6y^6z^6)
(x^3y^3 + y^3z^3 + z^3x^3)/3 ≥ ∛(1/27)
(x^3y^3 + y^3z^3 + z^3x^3)/3 ≥ 1/3
Multiplying both sides by 3:
x^3y^3 + y^3z^3 + z^3x^3 ≥ 1
So, the maximum possible value of x^3y^3 + y^3z^3 + z^3x^3 is 1.
Now, we can see that a=1 and b=1, since 1 = 1*1. Therefore, the value of a+b is 1+1=2.