1960 IMO Problems/Problem 2

Problem

For what values of the variable $x$ does the following inequality hold:

$\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?$

Solution

Set $x = -\frac{1}{2} + \frac{a^2}{2}$ , where $a\ge0$ . $\frac{4\left(-\frac{1}{2}+\frac{a^2}{2}\right)^2}{\left(1-\sqrt{1+2\left(-\frac{1}{2}+\frac{a^2}{2}\right)}\right)^2}<2\left(-\frac{1}{2}+\frac{a^2}{2}\right)+9$

After simplifying, we get $(a+1)^2<a^2+8$

So $a^2+2a+1<a^2+8$

Which gives $a<\frac{7}{2}$ and hence $-\frac{1}{2} \le x<\frac{45}{8}$ .

But $x=0$ makes the LHS indeterminate.

So, answer: $-\frac{1}{2} \le x<\frac{45}{8}$ , except $x=0$ .

Solution 2

If $x \neq 0$ , 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 $2x + 2\sqrt{2x + 1} + 2 < 2x + 9.$ or $\sqrt{2x + 1} < \frac{7}{2}.$

So $2x + 1 < 49/4$ and therefore $x < 45/8$ . But If $x < -1/2$ then the inequality makes no sense, since $\sqrt{2x + 1}$ is imaginary. So the original inequality holds if $x$ is in $[-1/2, 0) \cup (0, 45/8).$

1960 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1960 IMO Problems/Problem 2

Contents

Problem

Solution

Solution 2

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