在初中数学竞赛试题中,常常出现求最大值或最小值问题.除了利用二次函数的配方法求最值外,通常还借助于不等式“a+b≥2(?)”及一元二次方程的根的判别式“△”求解.灵活利用“a+b≥2(?)(a≥0,b≥0)”这个重要而基本的不等式,可以解决不少竞赛中棘手的最值问题.例1(2002年上海初中数学竞赛)若正数x、y、z满足xyz(x+y+z)=4,求(z+y)(y+z)的最小值.分析代数式(x+y)(y+z)展开整理可 In the junior high school mathematics contest questions, the problem of finding the maximum or minimum value often occurs. In addition to using the quadratic function matching method to find the maximum value, usually by means of the inequality “a+b≥2(?)” and the unary quadratic equation The root of the discriminant “△” Solve. Flexible use of “a + b ≥ 2 (?) (a ≥ 0, b ≥ 0)” this important and basic inequality, you can solve a lot of competition in the most troublesome problem. Example 1 (Shanghai Junior High School Math Competition 2002) If the positive x, y, z satisfies xyz(x+y+z)=4, find the minimum of (z+y)(y+z). Analyze the algebraic formula (x+ y)(y+z) Expand collation