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Algebra Difficulty 4.5 AIME Prove it Slovenia

Let xx, yy and zz be real numbers such that 0x,y,z10 \le x, y, z \le 1. Prove that
xyz+(1x)(1y)(1z)1. xyz + (1-x)(1-y)(1-z) \le 1.
When does the equality hold?

Solution

The inequality is equivalent to xy+yz+zxx+y+zxy + yz + zx \le x + y + z. Since xx is positive and y1y \le 1 we have xyxxy \le x. A similar reasoning shows that yzyyz \le y and zxzzx \le z. Hence, the inequality holds.

The equality holds if and only if xy=xxy = x, yz=yyz = y and zx=zzx = z. If x=0x=0, then the third equality implies z=0z=0 and the second equality then implies y=0y=0. Otherwise, we must have y=1y=1 and then the second equation implies z=1z=1 and from the third equation we get x=1x=1. Hence, the equality holds when x=y=z=0x=y=z=0 or x=y=z=1x=y=z=1.

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