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I have $3$ positive integers $i, k, q$, satisfying $1 \le k \le i \le q$. I am trying to prove the following inequality: $$\frac{{\left( {\begin{array}{*{20}{c}} {q - k}\\ {i - k} \end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}} q\\ i \end{array}} \right)}} = \frac{{\left( {i - k + 1} \right) \cdot \left( {i - k + 2} \right) \cdot ... \cdot \left( i \right)}}{{\left( {q - k + 1} \right) \cdot \left( {q - k + 2} \right) \cdot ... \cdot \left( q \right)}} \le \frac{{{i^k}}}{{{q^k}}}$$

Note that ${\left( {i - k + 1} \right) \cdot \left( {i - k + 2} \right) \cdot ... \cdot \left( i \right) \le {i^k}}$ and ${\left( {q - k + 1} \right) \cdot \left( {q - k + 2} \right) \cdot ... \cdot \left( q \right) \le {q^k}}$, making the inequality non-trivial. I considered the following cases:

  1. $k=1$. The inequality becomes $1 \le 1$ which is of course true.
  2. $k=2$. The inequality becomes $i \le q$ which is true.
  3. $k=3$. The inequality becomes $\frac{{q + i}}{{qi}} \le \frac{3}{2}$ which seems true, but proof is needed.

For $k=4$ and on the expressions become complicated (unless they can be simplified somehow). Any help would be highly appreciated.

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Hint. Show that $\frac{i-k+1}{q-k+1} \leq \frac i q$

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  • $\begingroup$ Done. Now by showing that $\frac{{i - k + j}}{{q - k + j}} \le \frac{i}{q}$ for $j \le k$ the inequality follows. Thanks. $\endgroup$ – r1c Feb 25 '14 at 18:28

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