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Problem 342

AMC 12 late, AIME early

Algebra Difficulty 4.7 Prove it Mediterranean Mathematical Competition · Greece

Let $x > 1$ a non integer number. Prove that
$\left( \frac{x+\{x\}}{[x]} - \frac{[x]}{x+\{x\}} \right) + \left( \frac{x+[x]}{\{x\}} - \frac{\{x\}}{x+[x]} \right) > \frac{9}{2},$
where $[x]$ and $\{x\}$ represents the integer and the fractional part of $x$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

We put $[x] = a$ , $\{x\} = r$ , where $0 \le r < 1$ . Then the given inequality becomes
$\left( \frac{a+2r}{a} - \frac{a}{a+2r} \right) + \left( \frac{2a+r}{r} - \frac{r}{2a+r} \right) > \frac{9}{2} \\ \Leftrightarrow 2 \left( \frac{r}{a} + \frac{a}{r} \right) - \left( \frac{a}{a+2r} + \frac{r}{2a+r} \right) > \frac{5}{2}.$
Since $\frac{r}{a} + \frac{a}{r} \ge 2$ , it is enough to prove that
$\frac{a}{a+2r} + \frac{r}{2a+r} < \frac{3}{2} \Leftrightarrow 0 < 2a^2 + 11ar + 2r^2 \Leftrightarrow 2(a+r)^2 + 7ar > 0,$
which is valid.

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