IMO 1998 SL 11

Origin: RUS

Problem

Let x, y, and z be positive real numbers such that xyz = 1. Prove that x3 (1 + y)(1 + z) + y3 (1 + z)(1 + x) + z3 (1 + x)(1 + y) \geq3 4.

Solution

The given inequality is equivalent to x3(x + 1) + y3(y + 1) + z3(z + 1) \geq 4(x + 1)(y + 1)(z + 1). By the A-G mean inequality, it will be enough to prove a stronger inequality: x4 + x3 + y4 + y3 + z4 + z3 \geq1 4[(x + 1)3 + (y + 1)3 + (z + 1)3]. (1) If we set Sk = xk+yk+zk, (1) takes the form S4+S3 \geq1 4S3+ 3 4S2+ 3 4S1+ 3 4. Note that by the A-G mean inequality, S1 = x+y+z \geq3. Thus it suffices to prove the following: If S1 \geq3 and m > n are positive integers, then Sm \geqSn. This can be shown in many ways. For example, by H¨older’s inequality, (xm + ym + zm)n/m(1 + 1 + 1)(m−n)/m \geqxn + yn + zn. (Another way is using the Chebyshev inequality: if x \geqy \geqz then xk−1 \geq yk−1 \geqzk−1; hence Sk = x \cdot xk−1 + y \cdot yk−1 + z \cdot zk−1 \geq1 3S1Sk−1, and the claim follows by induction.)

Second solution. Assume that x \geqy \geqz. Then also (y+1)(z+1) \geq (x+1)(z+1) \geq (x+1)(y+1). Hence Chebyshev’s inequality gives that x3 (1 + y)(1 + z) + y3 (1 + x)(1 + z) + z3 (1 + x)(1 + y) \geq1 (x3 + y3 + z3) \cdot (3 + x + y + z) (1 + x)(1 + y)(1 + z) . Now if we put x + y + z = 3S, we have x3 + y3 + z3 \geq3S and (1 + x)(1 + y)(1 + z) \leq(1 + a)3 by the A-G mean inequality. Thus the needed inequality reduces to 6S3 (1+S)3 \geq3 4, which is obviously true because S \geq1. Remark. Both these solutions use only that x + y + z \geq3.