2007 JBMO Problems/Problem 1
Contents
Problem
Let be positive real number such that . Prove that the equation has no real solution.
Solution
The discriminant of the equation is In order for the quadratic equation to have no real solution, the discriminant must be less than zero, so we need to show that That means we need to show that
Assume that Rearranging the equation results in If then would be negative, making the equality fail. If then , making However, that means so the equality also fails.
Thus, by proof by contradiction, must be greater than , so the discriminant of the equation is negative. That means the equation has no real solution.
Solution 2
Alternatively it is possible to show that or . However, the original expression also rearranges to . Therefore it is enough to prove that or . Evaluating and at and shows the intersection point is between these values, and there are clearly no more because the cube is convex on positive numbers. That means the quadratic in the question has no real solution.
See Also
2007 JBMO (Problems • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |