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数学报刊上介绍“一题多解”的文章不乏其例。本文涉及的一个无理不等式的几种巧证,没有沿袭有理化法证明无理不等式的固有模式(包括三角变换法化去根号的方法),旨在从开发发散性思维的角度,注重说明广泛而合理的联想对于拓宽解题思路、寻求最佳途径所起的作用。例已知f(x)=(1+x~2)~(1/2),a、b为相异实数,试证 |f(a)-f(b)|<|a-b|。证法1 欲证|f(a)-f(b)|<|a-b|,只须证|(f(a)-f(b))/(a-b)|<1。坐标平面上点A(a,f(a))与点B(b,f(b))决定的直线的斜率k_(AB)正是(f(a)-f(b))/(a-b),这样,原命题等价于证明|k_(AB)|<1,即是证明直线AB的倾斜角α小于45°或者大于135°。注意到f(x)=(1+x~2)~(1/2)的图象是双曲线y~2-x~2=1的上支,其渐近线
There are no shortages of examples in the articles on the “multiple explanations of one topic” published in the journals of several countries. In this paper, several irrational inequalities involved in several kinds of clever proofs, there is no follow rationalization method to prove the intrinsic mode of inequalities (including the method of triangulation transformation to remove the root number), designed to develop a divergence of thinking, pay attention to a wide range of Reasonable association plays a role in broadening the problem-solving ideas and seeking the best way. For example, it is known that f(x) = (1 + x ~ 2) ~ (1/2), a, b are different real numbers, test |f(a)-f(b)|<|a-b|. The proof method 1 wants to prove that |f(a)-f(b)|<|a-b| only requires |(f(a)-f(b))/(a-b)|<1. The slope k_(AB) of the line determined by the point A(a,f(a)) and the point B(b,f(b)) on the coordinate plane is exactly the same as (f(a)-f(b))/(ab) In this way, the original proposition is equivalent to the proof |k_(AB)|<1, which means that the inclination angle α of the straight line AB is less than 45° or greater than 135°. Notice that the image of f(x)=(1+x~2)~(1/2) is the upper branch of the hyperbola y~2-x~2=1 and its asymptote