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不等式的证明方法很多,有时使人觉得扑朔迷离,无从下手或证明太繁而通过联想构造函数,将常量作为变量的瞬时状态置于构造函数的定义域内,利用函数的性质证明不等式,却是十分巧妙有效的方法．本文介绍构造函数证明不等式的几种途径,读者可以体会到用函数思想证明不等式,思路清新、简捷明快．一、利用一次函数的保号性证明不等式例1 (第15届俄罗斯竞赛题)已知x,y,z ∈(0,1),求证:x(1-y)+y(1-z)+z(1-x) <1．
There are many ways to prove inequalities, and sometimes make people think complicated and confusing, unable to start or prove too complicated and through the associative constructor, the constants as variables in the transient state of the constructor domain, the use of the nature of the function to prove inequalities, but it is very clever effective method. This article describes several ways in which constructors prove inequalities. The reader can see that using inequality in function to prove inequalities, ideas are fresh, simple and neat. First, the use of a function of the degree of proof prove inequalities Example 1 (15th Russian competition) known x, y, z ∈ (0,1), verify: x (1-y) + y (1-z) + z (1-x) <1.