论文部分内容阅读
本文将给出不能直接运用二元平均值不等式处理的“积定和最小”一类问题的简捷处理模式.我们先看下述命题的证明过程.命题对于函数 f(x)=x+(a~2/x)(x∈R~+,a 为正常数),设 b 为正常数.(1)若 x≤ba,则函数 f(x)当 x≥b时的最小值是 [f(x)]_(min)=f(b).证 f(x)=(x+(b~2/x))+(a~2-b~2/x).(1)若 x≤bO,于是
In this paper, we will give a brief and simple processing model that can’t directly deal with the problems of “integral and minimal” which are handled by the binary mean inequality. Let’s look at the proof process of the following propositions. The proposition is for the function f(x)=x+( a~2/x)(x∈R~+,a is a normal number), let b be a normal number. (1) If x≤ba, then the minimum value of function f(x) when x≥b is [f(x)]_(min )=f(b). Proof f(x)=(x+(b~2/x))+(a~2-b~2/x). (1) If x≤bO, so