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Part of 1998 Moldova Team Selection Test

Problems(1)

$$\max_{1\leq i\leq n} |a_i-f(a_i)| \geq \max_{1\leq i\leq n} |a_i-b_i|$$

Source: Moldova TST 1998

8/8/2023
Let A={a1,a2,…,an}A=\{a_1,a_2,\ldots,a_n\} be a set with a1<a2…<ana_1<a_2\ldots<a_n and B={b1,b2,…,bn}B=\{b_1,b_2,\ldots,b_n\} be a set with b1<b2…<bnb_1<b_2\ldots<b_n. Show that for every bijective function f:A→Bf:A\rightarrow B the following relation takes place max⁡1≤i≤n∣ai−f(ai)∣≥max⁡1≤i≤n∣ai−bi∣.\max_{1\leq i\leq n} |a_i-f(a_i)| \geq \max_{1\leq i\leq n} |a_i-b_i|.
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