IMO 1970 SL 9

Origin: GDR

Problem

Let u1, u2, . . . , un, v1, v2, . . . , vn be real numbers. Prove that 1 + n  i=1 (ui + vi)2 \leq4 1 + n  i=1 u2 i 1 + n  i=1 v2 i . In what case does equality hold?

Solution

Let us set a = n i=1 u2 i and b = n i=1 v2 i . By Minkowski’s inequality (for p = 2) we have n i=1(ui + vi)2 \leq(a + b)2. Hence the LHS of the desired inequality is not greater than 1 + (a + b)2, while the RHS is equal to 4(1 + a2)(1 + b2)/3. Now it is sufficient to prove that 3 + 3(a + b)2 \leq4(1 + a2)(1 + b2). The last inequality can be reduced to the trivial 0 \leq(a −b)2 + (2ab −1)2. The equality in the initial inequality holds if and only if ui/vi = c for some c \inR and a = b = 1/ \sqrt 2.