IMO 1974 SL 4

Problem

I 4 (USS 4) The sum of the squares of five real numbers a1, a2, a3, a4, a5 equals 1. Prove that the least of the numbers (ai −aj)2, where i, j = 1, 2, 3, 4, 5 and i ̸= j, does not exceed 1/10.

Solution

Assume w.l.o.g. that a1 \leqa2 \leqa3 \leqa4 \leqa5. If m is the least value of |ai −aj|, i ̸= j, then ai+1 −ai \geqm for i = 1, 2, . . . , 5, and consequently ai −aj \geq(i −j)m for any i, j \in{1, . . ., 5}, i > j. Then it follows that  i>j (ai −aj)2 \geqm2  i>j (i −j)2 = 50m2. On the other hand, by the condition of the problem,  i>j (ai −aj)2 = 5  i=1 a2 i −(a1 + \cdot \cdot \cdot + a5)2 \leq5. Therefore 50m2 \leq5; i.e., m2 \leq 10.