If $a$ and $b$ are positive real numbers and each of the equations $x^2+3ax+b=0$ and $x^2+bx+3a=0$ has real roots, then the smallest value of $(a+b)$ is
- $\frac{16}{3}$
- $6$
- $\frac{14}{3}$
- $4$
For the roots to be real, the discriminant must be non-negative. For x^2 + 3ax + b = 0, D = 9a^2 - 4b >= 0. For x^2 + bx + 3a = 0, D = b^2 - 12a >= 0. We want to minimize a + b. Testing values that satisfy both inequalities, the minimum occurs at a = 4/3 and b = 4, giving a + b = 16/3.
For the equation x^2 + 3ax + b = 0 to have real roots, its discriminant must be non-negative, meaning (3a)^2 - 4b >= 0, or b <= 9a^2 / 4. Similarly, for the equation x^2 + bx + 3a = 0 to have real roots, the discriminant requires b^2 - 12a >= 0, or a <= b^2 / 12. Adding these two inequalities yields a new relationship, which after multiplying by 12 and rearranging gives 9a^2 - 16a + 4b^2 >= 0. To find the minimum possible value of a + b, we can add (a + b) to both sides and complete the square for the terms involving b. The expression 4b^2 + 4b + 1 is a perfect square, allowing the inequality to be rewritten as 9a^2 - 16a + 4(b + 1/2)^2 >= -1. Evaluating the boundary point of the sum a + b at the minimum condition of this quadratic reveals the smallest possible value to be 16/3.