设函数f﹙x﹚＝x²－1,对任意x∈[2／3,＋∞﹚,f﹙x／m﹚－4m²f﹙x﹚≤f﹙x－1﹚＋4f﹙m﹚恒成

把f(x)=x平方-1代入,得：x^2/m^2-1-4m^2(x^2-1)≤【（x-1)^2-1】+4（m^2-1)展开,消去4m^2,得：x^2/m^2-1-4m^2x^2≤x^2-2x-4把x^2项合并,常数合并,得：（1/m^2-4m^2-1)x^2≤-2x-3因为x≠0,所以1/m^2-4m^2-1≤（-2x-3)...