2004 IMO Problems/Problem 4

Problem

(Hojoo Lee) Let $n \geq 3$ be an integer. Let $t_1, t_2, \dots , t_n$ be positive real numbers such that

$n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).$

Show that $t_i$ , $t_j$ , $t_k$ are side lengths of a triangle for all $i$ , $j$ , $k$ with $1 \leq i < j < k \leq n$ .

Solution

For $n=3$ , suppose (for sake of contradiction) that $t_3 = t_2 + t_1 + k$ for $k \ge 0$ ; then (by Cauchy-Schwarz Inequality)

$\begin{align*}10 &> [2(t_1 + t_2) + k]\left(\frac {1}{t_1} + \frac {1}{t_2} + \frac 1{t_1 + t_2 + k}\right) = 2(t_1+t_2)\left(\frac 1{t_1} + \frac{1}{t_2}\right) + \left(\frac{k}{t_1} + \frac k{t_2}\right) + \frac{2t_1 + 2t_2 + k}{t_1 + t_2 + k}\\ &\ge 8 + \left(\frac{k}{t_1} + \frac k{t_2}\right) + 2 - \frac{k}{t_1 + t_2 + k}\\ &= 10 + k\left(\frac{1}{t_1} + \frac {1}{t_2} - \frac{1}{t_1 + t_2+k}\right) \ge 10\end{align*}$

so it is true for $n=3$ . We now claim the result by induction; for $n \ge 4$ , we have

$\begin{align*}f(n) &:= \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right) \\ &= \left( t_1 + t_2 + ... + t_{n-1} \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_{n-1}} \right) + t_n \sum_{i=1}^{n-1} \frac 1{t_i} + \frac{1}{t_n} \sum_{i=1}^{n-1} t_i + 1\end{align*}$

By AM-GM, $\frac{t_n}{t_i} + \frac{t_i}{t_n} \ge 2$ , so $f(n) \ge f(n-1) + 2(n-1) + 1 = f(n-1) + 2n - 1$ . Then the problem is reduced to proving the statement true for $n-1$ numbers, as desired. $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$

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2004 IMO Problems/Problem 4

Problem

Solution

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