均值不等式是解决最值的重要工具,但由于其约束条件苛刻,不少同学在使用时常常顾此失彼,导致解题失误.下面以同学们易陷于的误区举例分析如下:一、忽视等号成立条件例1求y=sinxcosx+sinx1cosx(0xπ)的最小值.错解:y=sinxcosx+sinx1cosx≥2sinxcosx·sinx1cosx=2.故ymin=2.剖 Mean value inequality is an important tool to solve the most value, but due to its harsh constraints, many students often overlooked each other in use, resulting in problem solving mistakes. The following examples of students mistakes easily fall into the following analysis: First, ignore the conditions for the establishment of equal signs Example 1 Find the minimum value of y=sinxcosx+sinx1cosx(0xπ). False solution: y=sinxcosx+sinx1cosx≥2sinxcosx·sinx1cosx=2. So ymin=2.