2008年第49届IMO试题里有这样一道不等式题:题目设实数x、y、z都不等于1,满足xyz=1,求证:x~2/((x-1)~2)+y~2/((y-1)~2)+z~2/((z-1)~2)≥1.(1)证明:设x/(x-1)=a,y/(y-1)=b,z/(z-1)=c,则x=a/(a-1),y=b/(b-1),z=c/(c-1).于是,有(abc)/((a-1)(b-1)(c-1))=1,化简,得ab+bc+ca=a+b+c-1.①显然,所要证明的不等式等价于a~2+b~2+c~2≥1.对该不等式的两边同时加上2(ab+bc+ca),便得(a+b+c)~2≥2(ab+bc+ca)+1.② There is such an inequality problem in the 49th IMO test paper in 2008. The title sets the real number x, y, and z are not equal to 1, and satisfies xyz=1. Proof: x~2/((x-1)~2)+y~ 2/((y-1)~2)+z~2/((z-1)~2)≥1.(1) Proof: Let x/(x-1)=a,y/(y-1 )=b,z/(z-1)=c, then x=a/(a-1), y=b/(b-1), z=c/(c-1). Then, there is (abc )/((a-1)(b-1)(c-1))=1, simplify and get ab+bc+ca=a+b+c-1.1 Obviously, the inequality to be proved is equivalent to a~2+b~2+c~2≥1. Add 2(ab+bc+ca) to both sides of the inequality to obtain (a+b+c)~2≥2(ab+bc+ca) )+1.2