Question

高中数学选修4-5不等式第二问求解 x+y＞0 xy≠0这个条件怎么用

高中数学选修4-5不等式第二问求解

x+y＞0 xy≠0这个条件怎么用

Answer 1

(1)
(x³+y³)-(x²y+y²x)
=(x+y)(x²-xy+y²)-xy(x+y)
=(x+y)(x²-2xy+y²)
=(x+y)(x-y)²
平方项恒非负，(x-y)²≥0，又已知x+y>0，因此(x+y)(x-y)²≥0
(x³+y³)-(x²y+y²x)≥0
x³+y³≥x²y+y²x，不等式成立。
(2)
xy≠0，x、y均不等于0，不等式有意义。
(x/y²+ y/x²)-(m/2)(1/x+1/y)
=(x³+y³)/(xy)² -(m/2)xy(x+y)/(xy)²
=[(x³+y³)-(m/2)xy(x+y)]/(xy)²
=[(x+y)(x²-xy+y²)-(m/2)xy(x+y)]/(xy)²
=(x+y)[x²+y²-(1+ m/2)xy]/(xy)²
(x/y²+ y/x²)≥(m/2)(1/x+1/y)恒成立，即(x+y)[x²+y²-(1+ m/2)xy]/(xy)²恒≥0
xy≠0，(xy)²恒>0，由已知得x+y>0，要不等式恒成立，只需x²+y²-(1+ m/2)xy恒≥0
[-(1+ m/2)]²-4≤0
-2≤1+ m/2≤2
解得-6≤m≤2
m的取值范围为[-6，2]