IMO 1993 SL 24

Origin: USA

Problem

Prove that a b + 2c + 3d + b c + 2d + 3a + c d + 2a + 3b + d a + 2b + 3c \geq2 for all positive real numbers a, b, c, d.

Solution

By the Cauchy–Schwarz inequality, if x1, x2, . . . , xn and y1, y2, . . . , yn are positive numbers, then

n  i=1 xi yi

n  i=1 xiyi

\geq

n  i=1 xi . Applying this to the numbers a, b, c, d and b+2c+3d, c+2d+3a, d+2a+ 3b, a + 2b + 3c (here n = 4), we obtain a b + 2c + 3d + b c + 2d + 3a + c d + 2a + 3b + d a + 2b + 3c \geq (a + b + c + d)2 4(ab + ac + ad + bc + bd + cd) \geq2 3. The last inequality follows, for example, from (a −b)2 + (a −c)2 + \cdot \cdot \cdot + (c −d)2 \geq0. Equality holds if and only if a = b = c = d. Second solution. Putting A = b+2c+3d, B = c+2d+3a, C = d+2a+3b, D = a + 2b + 3c, our inequality transforms into −5A + 7B + C + D 24A

• −5B + 7C + D + A 24B +−5C + 7D + A + B 24C
• −5D + 7A + B + C 24D \geq2 3 . This follows from the arithmetic-geometric mean inequality, since B A + C B + D C + A D \geq4, etc.