1990 APMO Problems/Problem 2

We have to prove $S_k S_{n-k} \geq {n \choose k}^2 a_1 a_2 \cdots a_n.$

Let, $\begin{aligned} S_{k} & = \sum_{i = 1}^{(\binom{n}{k})} t_{i} \\ S_{n - k} & = (\prod_{m = 1}^{n} a_{m}) (\sum_{i = 1}^{(\binom{n}{k})} \frac{1}{t_{i}}) \end{aligned}$ Thus, $\begin{aligned} S_{k} S_{n - k} & = (\sum_{i = 1}^{(\binom{n}{k})} t_{i}) (\prod_{m = 1}^{n} a_{m}) (\sum_{i = 1}^{(\binom{n}{k})} \frac{1}{t_{i}}) \\ = (\prod_{m = 1}^{n} a_{m}) [\sum_{i = 1}^{(\binom{n}{k})} 1 + {\sum_{i = 1}^{(\binom{n}{k})} \sum_{j = 1}^{(\binom{n}{k})}}_{i \neq j} \frac{t_{i}}{t_{j}}] \\ = (\prod_{m = 1}^{n} a_{m}) [(\binom{n}{k}) + \sum_{i, j} (\frac{t_{i}}{t_{j}} + \frac{t_{j}}{t_{i}})] \\ \geq (\prod_{m = 1}^{n} a_{m}) [(\binom{n}{k}) + 2 \cdot \frac{{(\binom{n}{k})}^{2} - (\binom{n}{k})}{2}] \\ = {(\binom{n}{k})}^{2} \prod_{m = 1}^{n} a_{m} ◻ \end{aligned}$

In the fourth line, we used the AM-GM inequality and used the fact that there are $\dfrac{1}{2}\left({n \choose k}^2-{n \choose k}\right)$ terms in the sum of third line.

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1990 APMO Problems/Problem 2