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 a/b where a and b are positive, coprime integers. What is the value of a+b?
To find the maximum value of x^3y^3 + y^3z^3 + z^3x^3, we can use the AM-GM inequality.
The AM-GM inequality states that for any non-negative real numbers x and y, the geometric mean is always less than or equal to the arithmetic mean.
In this case, we have three terms: x^3y^3, y^3z^3, and z^3x^3. We want to maximize their sum.
By applying the AM-GM inequality, we have:
(x^3y^3 + y^3z^3 + z^3x^3) / 3 ≥ (x^3y^3 * y^3z^3 * z^3x^3)^(1/3)
Simplifying, we get:
x^3y^3 + y^3z^3 + z^3x^3 ≥ 3∛(x^9y^9z^9)
Since x + y + z = 1, we can rewrite this as:
x + y + z = 1
(x + y + z)^9 = 1^9
Expanding the left side of the equation using the binomial theorem, we get:
x^9 + 9x^8y + 36x^7y^2 + 84x^6y^3 + 126x^5y^4 + 126x^4y^5 + 84x^3y^6 + 36x^2y^7 + 9xy^8 + y^9 + ...
Since we are looking for the maximum value of x^3y^3 + y^3z^3 + z^3x^3, we only need to consider the terms with powers of x^3, y^3, and z^3. The highest possible power of x^3, y^3, or z^3 is 9, so we only need to consider the terms up to that power.
Based on the expansion, we can observe that the maximum value of x^3y^3 + y^3z^3 + z^3x^3 occurs when x^3 = y^3 = z^3. This maximizes the sum.
Since x + y + z = 1, we can divide the equation by 3 to solve for x^3:
(x + y + z)/3 = (1/3)^3
x^3 = (1/3)^3
Similarly, y^3 = z^3 = (1/3)^3
So, x^3y^3 + y^3z^3 + z^3x^3 = (1/3)^6 + (1/3)^6 + (1/3)^6
Simplifying, we get:
x^3y^3 + y^3z^3 + z^3x^3 = 3 * (1/3)^6
x^3y^3 + y^3z^3 + z^3x^3 = 3/729
The maximum value of x^3y^3 + y^3z^3 + z^3x^3 is 3/729.
Therefore, a = 3 and b = 729.
So, the value of a + b is 3 + 729 = 732.