巧用基本不等式等号成立条件解题

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基本不等式:√ab≤a+b2(其中a≥0,b≥0),当且仅当a=b时等号成立.当√ab=a+b2时,即12(√a-√b)2=0,可看出a=b.a=b一方面可看作不等式成立的特殊情况,另一方面也可看作恒等式成立的条件.基本不等式等号成立的条件有两个:1两数非负,2两数相等,这就说明基本不等式等号成立对条件有着较强的要求.反过来,如果基本不等式的等号成立,我们能得到一些等式,利用这些等式我们可以解高次方 The basic inequality: √ab≤a + b2 (where a≥0, b≥0), and if and only if a = b the equal sign holds.When √ab = a + b2, that is 12 (√a-√b) 2 = 0, it can be seen that a = ba = b can be considered as a special case of inequality on the one hand and as a condition of equality established on the other. There are two conditions for the establishment of the basic inequality: Negative, 2 equal to two, which shows that the establishment of basic inequality has strong requirements on the conditions. Conversely, if the equality of basic inequalities holds, we can get some equations, we can use these equations to solve higher order square
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