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在数学竞赛中时有满足条件abc=k~3的不等式证明题出现.证明这样的不等式的一种处理方法是:令a=k·x/y,b=k·y/z,c=k·z/x等;特别地,当abc=1时,可令a=x/y,b=y/z,c=z/x等.以下通过举例说明这种方法.例1(第41届IMO第2题)设a,b,c为正数,且满足abc一1,证明:(a-1+1/b)(b-1+1/c)(c-1+1/a)≤1①证明由abc=1,令a=x/y,b=y/z,c=z/x(x,y,z>0),代入①,可知所证①式就是(x/y-1+z/y)(y/z-1+x/z)(z/x-1+y/x)≤1即(y+z一x)(z+x-y)(x+y-z)≤xyz②不等式②就是1983年瑞士数学竞赛试题,
In the mathematics competition, there is an inequality proving problem that satisfies the condition abc = k ~ 3. One way to prove such an inequality is to let a = k · x / y, b = k · y / z and c = k · Z / x and so on; in particular, when abc = 1, a = x / y, b = y / z, c = z / (A-1 + 1 / b) (b-1 + 1 / c) (c-1 + 1 / a) where a, b and c are positive numbers and satisfy abc- ≤1 ① Proof that abc = 1, let a = x / y, b = y / z, c = z / x (x, y, z> 0) 1 + z / y) (y / z-1 + x / z) (z / x-1 + y / x) ≤1 ie (y + z-x) (z + xy) Inequality ② is the 1983 Swiss Mathematical Contest test questions,