1960 IMO Problems/Problem 2
Contents
Problem
For what values of the variable does the following inequality hold:
Solution
Set , where .
After simplifying, we get
So
Which gives and hence .
But makes the LHS indeterminate.
So, answer: , except .
Solution 2
If , then the LHS is defined and rewrites as follows:
\begin{align*} \frac{4x^2}{(1 - \sqrt{2x + 1})^2} &= \biggl(\frac{2x}{1 - \sqrt{2x + 1}}\biggl)^2 \\ &= \biggl( \frac{2x}{1 - \sqrt{2x + 1}} \cdot \frac{1 + \sqrt{2x + 1}}{1 + \sqrt{2x + 1}} \biggl)^2 \\ &= (1 + \sqrt{2x + 1})^2 \\ &= 2x + 2\sqrt{2x + 1} + 2. \end{align*}
The inequality therefore holds if and only if or
So and therefore . But if then the inequality makes no sense, since is imaginary. So the original inequality holds iff is in
See Also
1960 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |