1965 IMO Problems/Problem 1
Problem
Determine all values in the interval which satisfy the inequality
Solution
We shall deal with the left side of the inequality first () and the right side after that.
It is clear that the left inequality is true when is non-positive, and that is when is in the interval . We shall now consider when is positive. We can square the given inequality, and the resulting inequality will be true whenever the original left inequality is true. . This inequality is equivalent to . I shall now divide this problem into cases.
Case 1: is non-negative. This means that is in one of the intervals or . We must find all in these two intervals such that . This inequality is equivalent to , which is only true when or .
Case 2: is negative. This means that is in one of the interavals or . We must find all in these two intervals such that , which is equivalent to , which is true for all in these intervals.
Therefore the left inequality is true when is in the union of the intervals , , and , which is the interval . We shall now deal with the right inequality.
As above, we can square it and have it be true whenever the original right inequality is true, so we do that. , which is always true. Therefore the original right inequality is always satisfied, and all such that the original inequality is satisfied are in the interval .
1965 IMO (Problems) • Resources) | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |