IMO 1972 SL 9

Origin: NET

Problem

Find all solutions in positive real numbers xi (i = 1, 2, 3, 4, 5) of the following system of inequalities: (x2 1 −x3x5)(x2 2 −x3x5) \leq0, (i) (x2 2 −x4x1)(x2 3 −x4x1) \leq0, (ii) (x2 3 −x5x2)(x2 4 −x5x2) \leq0, (iii) (x2 4 −x1x3)(x2 5 −x1x3) \leq0, (iv) (x2 5 −x2x4)(x2 1 −x2x4) \leq0. (v)

Solution

Clearly x1 = x2 = x3 = x4 = x5 is a solution. We shall show that this describes all solutions. Suppose that not all xi are equal. Then among x3, x5, x2, x4, x1 two con- secutive are distinct: Assume w.l.o.g. that x3 ̸= x5. Moreover, since (1/x1, . . . , 1/x5) is a solution whenever (x1, . . . , x5) is, we may assume that x3 < x5. Consider first the case x1 \leqx2. We infer from (i) that x1 \leq\sqrtx3x5 < x5 and x2 \geq\sqrtx3x5 > x3. Then x2 5 > x1x3, which together with (iv) gives x2 4 \leqx1x3 < x3x5; but we also have x2 3 \leqx5x2; hence by (iii), x2 4 \geq x5x2 > x5x3, a contradiction. Consider next the case x1 > x2. We infer from (i) that x1 \geq\sqrtx3x5 > x3 and x2 \leq\sqrtx3x5 < x5. Then by (ii) and (v), x1x4 \leqmax(x2 2, x2 3) \leqx3x5 and x2x4 \geqmin(x2 1, x2 5) \geqx3x5, which contradicts the assumption x1 > x2. Second solution. 0 \geqL1 = (x2 1 −x3x5)(x2 2 −x3x5) = x2 1x2 2 + x2 3x2 5 −(x2 1 + x2 2)x3x5 \geqx2 1x2 2 + x2 3x2 5 −1 2(x2 1x2 3 + x2 1x2 5 + x2 2x2 3 + x2 2x2 5), and analogously for L2, . . . , L5. Therefore L1 + L2 + L3 + L4 + L5 \geq0, with the only case of equality x1 = x2 = x3 = x4 = x5.